Calculus Examples

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

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.

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

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

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

Step 1.1.2.1.2

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

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

Step 1.1.2.3.2

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

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

Step 1.1.2.3.3

Multiply $- 1$ by $0$ .

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

Step 1.1.2.3.4

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

$\frac{0}{lim_{x \to π} 1 + \sec (x)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

$\frac{0}{lim_{x \to π} 1 + lim_{x \to π} \sec (x)}$

Step 1.1.3.1.2

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

$\frac{0}{1 + lim_{x \to π} \sec (x)}$

Step 1.1.3.1.3

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

$\frac{0}{1 + \sec (lim_{x \to π} x)}$

Step 1.1.3.2

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

$\frac{0}{1 + \sec (π)}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$\frac{0}{1 - \sec (0)}$

Step 1.1.3.3.1.2

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

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

Step 1.1.3.3.1.3

Multiply $- 1$ by $1$ .

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

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.4

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

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 $g (x) = \tan (x)$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $\tan (x)$ .

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

Step 1.3.2.2

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

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

Step 1.3.2.3

Replace all occurrences of $u$ with $\tan (x)$ .

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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

Step 1.3.6

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

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

Step 1.3.7

The derivative of $\sec (x)$ with respect to $x$ is $\sec (x) \tan (x)$ .

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

Step 1.3.8

Add $0$ and $\sec (x) \tan (x)$ .

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

Step 1.4

Reduce.

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Step 1.4.1

Cancel the common factor of $\sec^{2} (x)$ and $\sec (x)$ .

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Step 1.4.1.1

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

$lim_{x \to π} \frac{\sec (x) (2 \sec (x) \tan (x))}{\sec (x) \tan (x)}$

Step 1.4.1.2

Cancel the common factors.

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Step 1.4.1.2.1

Factor $\sec (x)$ out of $\sec (x) \tan (x)$ .

$lim_{x \to π} \frac{\sec (x) (2 \sec (x) \tan (x))}{\sec (x) (\tan (x))}$

Step 1.4.1.2.2

Cancel the common factor.

$lim_{x \to π} \frac{\sec (x) (2 \sec (x) \tan (x))}{\sec (x) \tan (x)}$

Step 1.4.1.2.3

Rewrite the expression.

$lim_{x \to π} \frac{2 \sec (x) \tan (x)}{\tan (x)}$

Step 1.4.2

Cancel the common factor of $\tan (x)$ .

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Step 1.4.2.1

Cancel the common factor.

$lim_{x \to π} \frac{2 \sec (x) \tan (x)}{\tan (x)}$

Step 1.4.2.2

Divide $2 \sec (x)$ by $1$ .

$lim_{x \to π} 2 \sec (x)$

Step 2

Evaluate the limit.

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Step 2.1

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

$2 lim_{x \to π} \sec (x)$

Step 2.2

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

$2 \sec (lim_{x \to π} x)$

Step 3

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

$2 \sec (π)$

Step 4

Simplify the answer.

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Step 4.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$2 (- \sec (0))$

Step 4.2

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

$2 (- 1 \cdot 1)$

Step 4.3

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

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Step 4.3.1

Multiply $- 1$ by $1$ .

$2 \cdot - 1$

Step 4.3.2

Multiply $2$ by $- 1$ .

$- 2$