Calculus Examples

$lim_{x \to 1} \frac{\tan^{2} (2 x - 2)}{x^{2} - 2 x + 1}$

Step 1

Apply L'Hospital's rule.

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Step 1.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 1.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 1} \tan^{2} (2 x - 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

Move the exponent $2$ from $\tan^{2} (2 x - 2)$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 1} \tan (2 x - 2))}^{2}}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.1.2

Move the limit inside the trig function because tangent is continuous.

$\frac{\tan^{2} (lim_{x \to 1} 2 x - 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.1.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{\tan^{2} (lim_{x \to 1} 2 x - lim_{x \to 1} 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.1.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{\tan^{2} (2 lim_{x \to 1} x - lim_{x \to 1} 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.1.5

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$\frac{\tan^{2} (2 lim_{x \to 1} x - 1 \cdot 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{\tan^{2} (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Simplify each term.

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Step 1.1.2.3.1.1

Multiply $2$ by $1$ .

$\frac{\tan^{2} (2 - 1 \cdot 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.3.1.2

Multiply $- 1$ by $2$ .

$\frac{\tan^{2} (2 - 2)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.3.2

Subtract $2$ from $2$ .

$\frac{\tan^{2} (0)}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.3.3

The exact value of $\tan (0)$ is $0$ .

$\frac{0^{2}}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.2.3.4

Raising $0$ to any positive power yields $0$ .

$\frac{0}{lim_{x \to 1} x^{2} - 2 x + 1}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$\frac{0}{lim_{x \to 1} x^{2} - lim_{x \to 1} 2 x + lim_{x \to 1} 1}$

Step 1.1.3.2

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{0}{{(lim_{x \to 1} x)}^{2} - lim_{x \to 1} 2 x + lim_{x \to 1} 1}$

Step 1.1.3.3

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{{(lim_{x \to 1} x)}^{2} - 2 lim_{x \to 1} x + lim_{x \to 1} 1}$

Step 1.1.3.4

Evaluate the limit of $1$ which is constant as $x$ approaches $1$ .

$\frac{0}{{(lim_{x \to 1} x)}^{2} - 2 lim_{x \to 1} x + 1}$

Step 1.1.3.5

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 1.1.3.5.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{0}{1^{2} - 2 lim_{x \to 1} x + 1}$

Step 1.1.3.5.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$\frac{0}{1^{2} - 2 \cdot 1 + 1}$

Step 1.1.3.6

Simplify the answer.

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Step 1.1.3.6.1

Simplify each term.

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Step 1.1.3.6.1.1

One to any power is one.

$\frac{0}{1 - 2 \cdot 1 + 1}$

Step 1.1.3.6.1.2

Multiply $- 2$ by $1$ .

$\frac{0}{1 - 2 + 1}$

Step 1.1.3.6.2

Subtract $2$ from $1$ .

$\frac{0}{- 1 + 1}$

Step 1.1.3.6.3

Add $- 1$ and $1$ .

$\frac{0}{0}$

Step 1.1.3.6.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.1.3.7

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 1} \frac{\tan^{2} (2 x - 2)}{x^{2} - 2 x + 1} = lim_{x \to 1} \frac{\frac{d}{d x} [\tan^{2} (2 x - 2)]}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 1} \frac{\frac{d}{d x} [\tan^{2} (2 x - 2)]}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \tan (2 x - 2)$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u_{1}$ as $\tan (2 x - 2)$ .

$lim_{x \to 1} \frac{\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\tan (2 x - 2)]}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$lim_{x \to 1} \frac{2 u_{1} \frac{d}{d x} [\tan (2 x - 2)]}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.2.3

Replace all occurrences of $u_{1}$ with $\tan (2 x - 2)$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \frac{d}{d x} [\tan (2 x - 2)]}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x - 2$ .

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Step 1.3.3.1

To apply the Chain Rule, set $u_{2}$ as $2 x - 2$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) (\frac{d}{d u_{2}} [\tan (u_{2})] \frac{d}{d x} [2 x - 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.3.2

The derivative of $\tan (u_{2})$ with respect to $u_{2}$ is $\sec^{2} (u_{2})$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) (\sec^{2} (u_{2}) \frac{d}{d x} [2 x - 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.3.3

Replace all occurrences of $u_{2}$ with $2 x - 2$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) (\sec^{2} (2 x - 2) \frac{d}{d x} [2 x - 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.4

By the Sum Rule, the derivative of $2 x - 2$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2]$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.5

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) (2 \cdot 1 + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.7

Multiply $2$ by $1$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) (2 + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.8

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) (2 + 0)}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.9

Add $2$ and $0$ .

$lim_{x \to 1} \frac{2 \tan (2 x - 2) \sec^{2} (2 x - 2) \cdot 2}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.10

Multiply $2$ by $2$ .

$lim_{x \to 1} \frac{4 \tan (2 x - 2) \sec^{2} (2 x - 2)}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.11

Reorder the factors of $4 \tan (2 x - 2) \sec^{2} (2 x - 2)$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{\frac{d}{d x} [x^{2} - 2 x + 1]}$

Step 1.3.12

By the Sum Rule, the derivative of $x^{2} - 2 x + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]}$

Step 1.3.13

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]}$

Step 1.3.14

Evaluate $\frac{d}{d x} [- 2 x]$ .

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Step 1.3.14.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [1]}$

Step 1.3.14.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x - 2 \cdot 1 + \frac{d}{d x} [1]}$

Step 1.3.14.3

Multiply $- 2$ by $1$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x - 2 + \frac{d}{d x} [1]}$

Step 1.3.15

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x - 2 + 0}$

Step 1.3.16

Add $2 x - 2$ and $0$ .

$lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan (2 x - 2)}{2 x - 2}$

Step 1.4

Cancel the common factor of $4$ and $2 x - 2$ .

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Step 1.4.1

Factor $2$ out of $4 \sec^{2} (2 x - 2) \tan (2 x - 2)$ .

$lim_{x \to 1} \frac{2 (2 \sec^{2} (2 x - 2) \tan (2 x - 2))}{2 x - 2}$

Step 1.4.2

Cancel the common factors.

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Step 1.4.2.1

Factor $2$ out of $2 x$ .

$lim_{x \to 1} \frac{2 (2 \sec^{2} (2 x - 2) \tan (2 x - 2))}{2 (x) - 2}$

Step 1.4.2.2

Factor $2$ out of $- 2$ .

$lim_{x \to 1} \frac{2 (2 \sec^{2} (2 x - 2) \tan (2 x - 2))}{2 (x) + 2 (- 1)}$

Step 1.4.2.3

Factor $2$ out of $2 (x) + 2 (- 1)$ .

$lim_{x \to 1} \frac{2 (2 \sec^{2} (2 x - 2) \tan (2 x - 2))}{2 (x - 1)}$

Step 1.4.2.4

Cancel the common factor.

$lim_{x \to 1} \frac{2 (2 \sec^{2} (2 x - 2) \tan (2 x - 2))}{2 (x - 1)}$

Step 1.4.2.5

Rewrite the expression.

$lim_{x \to 1} \frac{2 \sec^{2} (2 x - 2) \tan (2 x - 2)}{x - 1}$

Step 2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \tan (2 x - 2)}{x - 1}$

Step 3

Apply L'Hospital's rule.

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Step 3.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 3.1.1

Take the limit of the numerator and the limit of the denominator.

$2 \frac{lim_{x \to 1} \sec^{2} (2 x - 2) \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

Split the limit using the Product of Limits Rule on the limit as $x$ approaches $1$ .

$2 \frac{lim_{x \to 1} \sec^{2} (2 x - 2) \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.2

Move the exponent $2$ from $\sec^{2} (2 x - 2)$ outside the limit using the Limits Power Rule.

$2 \frac{{(lim_{x \to 1} \sec (2 x - 2))}^{2} \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.3

Move the limit inside the trig function because secant is continuous.

$2 \frac{\sec^{2} (lim_{x \to 1} 2 x - 2) \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.4

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \frac{\sec^{2} (lim_{x \to 1} 2 x - lim_{x \to 1} 2) \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.5

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - lim_{x \to 1} 2) \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.6

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - 1 \cdot 2) \cdot lim_{x \to 1} \tan (2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.7

Move the limit inside the trig function because tangent is continuous.

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - 1 \cdot 2) \cdot \tan (lim_{x \to 1} 2 x - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.8

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - 1 \cdot 2) \cdot \tan (lim_{x \to 1} 2 x - lim_{x \to 1} 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.9

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - 1 \cdot 2) \cdot \tan (2 lim_{x \to 1} x - lim_{x \to 1} 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.10

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$2 \frac{\sec^{2} (2 lim_{x \to 1} x - 1 \cdot 2) \cdot \tan (2 lim_{x \to 1} x - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.11

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 3.1.2.11.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$2 \frac{\sec^{2} (2 \cdot 1 - 1 \cdot 2) \cdot \tan (2 lim_{x \to 1} x - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.11.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$2 \frac{\sec^{2} (2 \cdot 1 - 1 \cdot 2) \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12

Simplify the answer.

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Step 3.1.2.12.1

Simplify each term.

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Step 3.1.2.12.1.1

Multiply $2$ by $1$ .

$2 \frac{\sec^{2} (2 - 1 \cdot 2) \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.1.2

Multiply $- 1$ by $2$ .

$2 \frac{\sec^{2} (2 - 2) \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.2

Subtract $2$ from $2$ .

$2 \frac{\sec^{2} (0) \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.3

The exact value of $\sec (0)$ is $1$ .

$2 \frac{1^{2} \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.4

One to any power is one.

$2 \frac{1 \cdot \tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.5

Multiply $\tan (2 \cdot 1 - 1 \cdot 2)$ by $1$ .

$2 \frac{\tan (2 \cdot 1 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.6

Simplify each term.

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Step 3.1.2.12.6.1

Multiply $2$ by $1$ .

$2 \frac{\tan (2 - 1 \cdot 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.6.2

Multiply $- 1$ by $2$ .

$2 \frac{\tan (2 - 2)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.7

Subtract $2$ from $2$ .

$2 \frac{\tan (0)}{lim_{x \to 1} x - 1}$

Step 3.1.2.12.8

The exact value of $\tan (0)$ is $0$ .

$2 \frac{0}{lim_{x \to 1} x - 1}$

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

Evaluate the limit.

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Step 3.1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \frac{0}{lim_{x \to 1} x - lim_{x \to 1} 1}$

Step 3.1.3.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $1$ .

$2 \frac{0}{lim_{x \to 1} x - 1 \cdot 1}$

Step 3.1.3.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$2 \frac{0}{1 - 1 \cdot 1}$

Step 3.1.3.3

Simplify the answer.

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Step 3.1.3.3.1

Multiply $- 1$ by $1$ .

$2 \frac{0}{1 - 1}$

Step 3.1.3.3.2

Subtract $1$ from $1$ .

$2 (\frac{0}{0})$

Step 3.1.3.3.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$2 (\frac{0}{0})$

Step 3.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$2 (\frac{0}{0})$

Step 3.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$2 (\frac{0}{0})$

Step 3.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \tan (2 x - 2)}{x - 1} = lim_{x \to 1} \frac{\frac{d}{d x} [\sec^{2} (2 x - 2) \tan (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$2 lim_{x \to 1} \frac{\frac{d}{d x} [\sec^{2} (2 x - 2) \tan (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \sec^{2} (2 x - 2)$ and $g (x) = \tan (2 x - 2)$ .

$2 lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \frac{d}{d x} [\tan (2 x - 2)] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 2 x - 2$ .

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Step 3.3.3.1

To apply the Chain Rule, set $u_{1}$ as $2 x - 2$ .

$2 lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \frac{d}{d u_{1}} [\tan (u_{1})] \frac{d}{d x} [2 x - 2] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.3.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

$2 lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \sec^{2} (u_{1}) \frac{d}{d x} [2 x - 2] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.3.3

Replace all occurrences of $u_{1}$ with $2 x - 2$ .

$2 lim_{x \to 1} \frac{\sec^{2} (2 x - 2) \sec^{2} (2 x - 2) \frac{d}{d x} [2 x - 2] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.4

Multiply $\sec^{2} (2 x - 2)$ by $\sec^{2} (2 x - 2)$ by adding the exponents.

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Step 3.3.4.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 lim_{x \to 1} \frac{{\sec (2 x - 2)}^{2 + 2} \frac{d}{d x} [2 x - 2] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.4.2

Add $2$ and $2$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) \frac{d}{d x} [2 x - 2] + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.5

By the Sum Rule, the derivative of $2 x - 2$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2]$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2]) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.6

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2]) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.7

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) (2 \cdot 1 + \frac{d}{d x} [- 2]) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.8

Multiply $2$ by $1$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) (2 + \frac{d}{d x} [- 2]) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.9

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) (2 + 0) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.10

Add $2$ and $0$ .

$2 lim_{x \to 1} \frac{\sec^{4} (2 x - 2) \cdot 2 + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.11

Move $2$ to the left of $\sec^{4} (2 x - 2)$ .

$2 lim_{x \to 1} \frac{2 \cdot \sec^{4} (2 x - 2) + \tan (2 x - 2) \frac{d}{d x} [\sec^{2} (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.12

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sec (2 x - 2)$ .

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Step 3.3.12.1

To apply the Chain Rule, set $u_{2}$ as $\sec (2 x - 2)$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + \tan (2 x - 2) \frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [\sec (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.12.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + \tan (2 x - 2) \cdot 2 u_{2} \frac{d}{d x} [\sec (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.12.3

Replace all occurrences of $u_{2}$ with $\sec (2 x - 2)$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + \tan (2 x - 2) \cdot 2 \sec (2 x - 2) \frac{d}{d x} [\sec (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.13

Move $2$ to the left of $\tan (2 x - 2)$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \cdot \tan (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [\sec (2 x - 2)]}{\frac{d}{d x} [x - 1]}$

Step 3.3.14

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 2 x - 2$ .

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Step 3.3.14.1

To apply the Chain Rule, set $u_{3}$ as $2 x - 2$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan (2 x - 2) \sec (2 x - 2) \frac{d}{d u_{3}} [\sec (u_{3})] \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.14.2

The derivative of $\sec (u_{3})$ with respect to $u_{3}$ is $\sec (u_{3}) \tan (u_{3})$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan (2 x - 2) \sec (2 x - 2) \sec (u_{3}) \tan (u_{3}) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.14.3

Replace all occurrences of $u_{3}$ with $2 x - 2$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan (2 x - 2) \sec (2 x - 2) \sec (2 x - 2) \tan (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.15

Raise $\tan (2 x - 2)$ to the power of $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{1} (2 x - 2) \tan (2 x - 2) \sec (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.16

Raise $\tan (2 x - 2)$ to the power of $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{1} (2 x - 2) \tan^{1} (2 x - 2) \sec (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.17

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 {\tan (2 x - 2)}^{1 + 1} \sec (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.18

Add $1$ and $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.19

Raise $\sec (2 x - 2)$ to the power of $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{1} (2 x - 2) \sec (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.20

Raise $\sec (2 x - 2)$ to the power of $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{1} (2 x - 2) \sec^{1} (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.21

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) {\sec (2 x - 2)}^{1 + 1} \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.22

Add $1$ and $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) \frac{d}{d x} [2 x - 2]}{\frac{d}{d x} [x - 1]}$

Step 3.3.23

By the Sum Rule, the derivative of $2 x - 2$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2]$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x - 1]}$

Step 3.3.24

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x - 1]}$

Step 3.3.25

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) (2 \cdot 1 + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x - 1]}$

Step 3.3.26

Multiply $2$ by $1$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) (2 + \frac{d}{d x} [- 2])}{\frac{d}{d x} [x - 1]}$

Step 3.3.27

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) (2 + 0)}{\frac{d}{d x} [x - 1]}$

Step 3.3.28

Add $2$ and $0$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 2 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2) \cdot 2}{\frac{d}{d x} [x - 1]}$

Step 3.3.29

Multiply $2$ by $2$ .

$2 lim_{x \to 1} \frac{2 \sec^{4} (2 x - 2) + 4 \tan^{2} (2 x - 2) \sec^{2} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30

Simplify.

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Step 3.3.30.1

Reorder terms.

$2 lim_{x \to 1} \frac{4 \sec^{2} (2 x - 2) \tan^{2} (2 x - 2) + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2

Simplify each term.

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Step 3.3.30.2.1

Rewrite $\sec (2 x - 2)$ in terms of sines and cosines.

$2 lim_{x \to 1} \frac{4 {(\frac{1}{\cos (2 x - 2)})}^{2} \tan^{2} (2 x - 2) + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.2

Apply the product rule to $\frac{1}{\cos (2 x - 2)}$ .

$2 lim_{x \to 1} \frac{4 \frac{1^{2}}{\cos^{2} (2 x - 2)} \tan^{2} (2 x - 2) + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.3

One to any power is one.

$2 lim_{x \to 1} \frac{4 \frac{1}{\cos^{2} (2 x - 2)} \tan^{2} (2 x - 2) + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.4

Combine $4$ and $\frac{1}{\cos^{2} (2 x - 2)}$ .

$2 lim_{x \to 1} \frac{\frac{4}{\cos^{2} (2 x - 2)} \tan^{2} (2 x - 2) + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.5

Rewrite $\tan (2 x - 2)$ in terms of sines and cosines.

$2 lim_{x \to 1} \frac{\frac{4}{\cos^{2} (2 x - 2)} {(\frac{\sin (2 x - 2)}{\cos (2 x - 2)})}^{2} + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.6

Apply the product rule to $\frac{\sin (2 x - 2)}{\cos (2 x - 2)}$ .

$2 lim_{x \to 1} \frac{\frac{4}{\cos^{2} (2 x - 2)} \cdot \frac{\sin^{2} (2 x - 2)}{\cos^{2} (2 x - 2)} + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.7

Combine.

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{2} (2 x - 2) \cos^{2} (2 x - 2)} + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.8

Multiply $\cos^{2} (2 x - 2)$ by $\cos^{2} (2 x - 2)$ by adding the exponents.

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Step 3.3.30.2.8.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{{\cos (2 x - 2)}^{2 + 2}} + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.8.2

Add $2$ and $2$ .

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{4} (2 x - 2)} + 2 \sec^{4} (2 x - 2)}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.9

Rewrite $\sec (2 x - 2)$ in terms of sines and cosines.

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{4} (2 x - 2)} + 2 {(\frac{1}{\cos (2 x - 2)})}^{4}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.10

Apply the product rule to $\frac{1}{\cos (2 x - 2)}$ .

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{4} (2 x - 2)} + 2 \frac{1^{4}}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.11

One to any power is one.

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{4} (2 x - 2)} + 2 \frac{1}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.2.12

Combine $2$ and $\frac{1}{\cos^{4} (2 x - 2)}$ .

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2)}{\cos^{4} (2 x - 2)} + \frac{2}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.3

Combine the numerators over the common denominator.

$2 lim_{x \to 1} \frac{\frac{4 \sin^{2} (2 x - 2) + 2}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.4

Factor $2$ out of $4 \sin^{2} (2 x - 2) + 2$ .

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Step 3.3.30.4.1

Factor $2$ out of $4 \sin^{2} (2 x - 2)$ .

$2 lim_{x \to 1} \frac{\frac{2 \cdot 2 \sin^{2} (2 x - 2) + 2}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.4.2

Factor $2$ out of $2$ .

$2 lim_{x \to 1} \frac{\frac{2 \cdot 2 \sin^{2} (2 x - 2) + 2 (1)}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.30.4.3

Factor $2$ out of $2 (2 \sin^{2} (2 x - 2)) + 2 (1)$ .

$2 lim_{x \to 1} \frac{\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x - 1]}$

Step 3.3.31

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$2 lim_{x \to 1} \frac{\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}}{\frac{d}{d x} [x] + \frac{d}{d x} [- 1]}$

Step 3.3.32

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$2 lim_{x \to 1} \frac{\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}}{1 + \frac{d}{d x} [- 1]}$

Step 3.3.33

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$2 lim_{x \to 1} \frac{\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}}{1 + 0}$

Step 3.3.34

Add $1$ and $0$ .

$2 lim_{x \to 1} \frac{\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}}{1}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$2 lim_{x \to 1} \frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)} \cdot 1$

Step 3.5

Multiply $\frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}$ by $1$ .

$2 lim_{x \to 1} \frac{2 (2 \sin^{2} (2 x - 2) + 1)}{\cos^{4} (2 x - 2)}$

Step 4

Evaluate the limit.

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Step 4.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot 2 lim_{x \to 1} \frac{2 \sin^{2} (2 x - 2) + 1}{\cos^{4} (2 x - 2)}$

Step 4.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $1$ .

$2 \cdot 2 \frac{lim_{x \to 1} 2 \sin^{2} (2 x - 2) + 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.3

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \cdot 2 \frac{lim_{x \to 1} 2 \sin^{2} (2 x - 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.4

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot 2 \frac{2 lim_{x \to 1} \sin^{2} (2 x - 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.5

Move the exponent $2$ from $\sin^{2} (2 x - 2)$ outside the limit using the Limits Power Rule.

$2 \cdot 2 \frac{2 {(lim_{x \to 1} \sin (2 x - 2))}^{2} + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.6

Move the limit inside the trig function because sine is continuous.

$2 \cdot 2 \frac{2 \sin^{2} (lim_{x \to 1} 2 x - 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.7

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \cdot 2 \frac{2 \sin^{2} (lim_{x \to 1} 2 x - lim_{x \to 1} 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.8

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - lim_{x \to 1} 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.9

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + lim_{x \to 1} 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.10

Evaluate the limit of $1$ which is constant as $x$ approaches $1$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{lim_{x \to 1} \cos^{4} (2 x - 2)}$

Step 4.11

Move the exponent $4$ from $\cos^{4} (2 x - 2)$ outside the limit using the Limits Power Rule.

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{{(lim_{x \to 1} \cos (2 x - 2))}^{4}}$

Step 4.12

Move the limit inside the trig function because cosine is continuous.

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{\cos^{4} (lim_{x \to 1} 2 x - 2)}$

Step 4.13

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $1$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{\cos^{4} (lim_{x \to 1} 2 x - lim_{x \to 1} 2)}$

Step 4.14

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{\cos^{4} (2 lim_{x \to 1} x - lim_{x \to 1} 2)}$

Step 4.15

Evaluate the limit of $2$ which is constant as $x$ approaches $1$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 lim_{x \to 1} x - 1 \cdot 2) + 1}{\cos^{4} (2 lim_{x \to 1} x - 1 \cdot 2)}$

Step 5

Evaluate the limits by plugging in $1$ for all occurrences of $x$ .

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Step 5.1

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 \cdot 1 - 1 \cdot 2) + 1}{\cos^{4} (2 lim_{x \to 1} x - 1 \cdot 2)}$

Step 5.2

Evaluate the limit of $x$ by plugging in $1$ for $x$ .

$2 \cdot 2 \frac{2 \sin^{2} (2 \cdot 1 - 1 \cdot 2) + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6

Simplify the answer.

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Step 6.1

Multiply $2$ by $2$ .

$4 \frac{2 \sin^{2} (2 \cdot 1 - 1 \cdot 2) + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2

Simplify the numerator.

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $2$ by $1$ .

$4 \frac{2 \sin^{2} (2 - 1 \cdot 2) + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.1.2

Multiply $- 1$ by $2$ .

$4 \frac{2 \sin^{2} (2 - 2) + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.2

Subtract $2$ from $2$ .

$4 \frac{2 \sin^{2} (0) + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.3

The exact value of $\sin (0)$ is $0$ .

$4 \frac{2 \cdot 0^{2} + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.4

Raising $0$ to any positive power yields $0$ .

$4 \frac{2 \cdot 0 + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.5

Multiply $2$ by $0$ .

$4 \frac{0 + 1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.2.6

Add $0$ and $1$ .

$4 \frac{1}{\cos^{4} (2 \cdot 1 - 1 \cdot 2)}$

Step 6.3

Simplify the denominator.

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Step 6.3.1

Simplify each term.

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Step 6.3.1.1

Multiply $2$ by $1$ .

$4 \frac{1}{\cos^{4} (2 - 1 \cdot 2)}$

Step 6.3.1.2

Multiply $- 1$ by $2$ .

$4 \frac{1}{\cos^{4} (2 - 2)}$

Step 6.3.2

Subtract $2$ from $2$ .

$4 \frac{1}{\cos^{4} (0)}$

Step 6.3.3

The exact value of $\cos (0)$ is $1$ .

$4 \frac{1}{1^{4}}$

Step 6.3.4

One to any power is one.

$4 (\frac{1}{1})$

Step 6.4

Cancel the common factor of $1$ .

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Step 6.4.1

Cancel the common factor.

$4 (\frac{1}{1})$

Step 6.4.2

Rewrite the expression.

$4 \cdot 1$

Step 6.5

Multiply $4$ by $1$ .

$4$