Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

Evaluate the limit.

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Step 1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{π}{2}$ .

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

Step 1.2.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{π}{2}$ .

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

Step 1.2.1.3

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

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

Step 1.2.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

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

Step 1.2.3

Simplify the answer.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

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

Step 1.2.3.1.2

Multiply $- 1$ by $1$ .

$\frac{1 - 1}{lim_{x \to \frac{π}{2}} 1 + \cos (2 x)}$

Step 1.2.3.2

Subtract $1$ from $1$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

Evaluate the limit.

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Step 1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{π}{2}$ .

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

Step 1.3.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{π}{2}$ .

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

Step 1.3.1.3

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

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

Step 1.3.1.4

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

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

Step 1.3.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

$\frac{0}{1 + \cos (2 \frac{π}{2})}$

Step 1.3.3

Simplify the answer.

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Step 1.3.3.1

Simplify each term.

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Step 1.3.3.1.1

Cancel the common factor of $2$ .

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Step 1.3.3.1.1.1

Cancel the common factor.

$\frac{0}{1 + \cos (2 \frac{π}{2})}$

Step 1.3.3.1.1.2

Rewrite the expression.

$\frac{0}{1 + \cos (π)}$

Step 1.3.3.1.2

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

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

Step 1.3.3.1.3

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

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

Step 1.3.3.1.4

Multiply $- 1$ by $1$ .

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

Step 1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 1.3.4

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

Undefined

$\frac{0}{0}$

Step 1.4

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

Undefined

$\frac{0}{0}$

Step 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

$lim_{x \to \frac{π}{2}} \frac{0 + \frac{d}{d x} [- \sin (x)]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 3.4

Evaluate $\frac{d}{d x} [- \sin (x)]$ .

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Step 3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

$lim_{x \to \frac{π}{2}} \frac{0 - \frac{d}{d x} [\sin (x)]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 3.4.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$lim_{x \to \frac{π}{2}} \frac{0 - \cos (x)}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 3.5

Subtract $\cos (x)$ from $0$ .

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

Step 3.6

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

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

Step 3.7

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

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 + \frac{d}{d x} [\cos (2 x)]}$

Step 3.8

Evaluate $\frac{d}{d x} [\cos (2 x)]$ .

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Step 3.8.1

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

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Step 3.8.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]}$

Step 3.8.1.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - \sin (u) \frac{d}{d x} [2 x]}$

Step 3.8.1.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - \sin (2 x) \frac{d}{d x} [2 x]}$

Step 3.8.2

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

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - \sin (2 x) (2 \frac{d}{d x} [x])}$

Step 3.8.3

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

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - \sin (2 x) (2 \cdot 1)}$

Step 3.8.4

Multiply $2$ by $1$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - \sin (2 x) \cdot 2}$

Step 3.8.5

Multiply $2$ by $- 1$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{0 - 2 \sin (2 x)}$

Step 3.9

Subtract $2 \sin (2 x)$ from $0$ .

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{- 2 \sin (2 x)}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

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

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Step 4.1.1

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

$\frac{lim_{x \to \frac{π}{2}} - \cos (x)}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Evaluate the limit.

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Step 4.1.2.1.1

Move the term $- 1$ outside of the limit because it is constant with respect to $x$ .

$\frac{- lim_{x \to \frac{π}{2}} \cos (x)}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.2.1.2

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

$\frac{- \cos (lim_{x \to \frac{π}{2}} x)}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.2.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

$\frac{- \cos (\frac{π}{2})}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.2.3

Simplify the answer.

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Step 4.1.2.3.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$\frac{- 0}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.2.3.2

Multiply $- 1$ by $0$ .

$\frac{0}{lim_{x \to \frac{π}{2}} - 2 \sin (2 x)}$

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

Evaluate the limit.

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Step 4.1.3.1.1

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

$\frac{0}{- 2 lim_{x \to \frac{π}{2}} \sin (2 x)}$

Step 4.1.3.1.2

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

$\frac{0}{- 2 \sin (lim_{x \to \frac{π}{2}} 2 x)}$

Step 4.1.3.1.3

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

$\frac{0}{- 2 \sin (2 lim_{x \to \frac{π}{2}} x)}$

Step 4.1.3.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

$\frac{0}{- 2 \sin (2 \frac{π}{2})}$

Step 4.1.3.3

Simplify the answer.

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Step 4.1.3.3.1

Cancel the common factor of $2$ .

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Step 4.1.3.3.1.1

Cancel the common factor.

$\frac{0}{- 2 \sin (2 \frac{π}{2})}$

Step 4.1.3.3.1.2

Rewrite the expression.

$\frac{0}{- 2 \sin (π)}$

Step 4.1.3.3.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 4.1.3.3.3

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

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

Step 4.1.3.3.4

Multiply $- 2$ by $0$ .

$\frac{0}{0}$

Step 4.1.3.3.5

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

Undefined

$\frac{0}{0}$

Step 4.1.3.4

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

Undefined

$\frac{0}{0}$

Step 4.1.4

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

Undefined

$\frac{0}{0}$

Step 4.2

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.

$lim_{x \to \frac{π}{2}} \frac{- \cos (x)}{- 2 \sin (2 x)} = lim_{x \to \frac{π}{2}} \frac{\frac{d}{d x} [- \cos (x)]}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$lim_{x \to \frac{π}{2}} \frac{\frac{d}{d x} [- \cos (x)]}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

$lim_{x \to \frac{π}{2}} \frac{- \frac{d}{d x} [\cos (x)]}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3.3

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$lim_{x \to \frac{π}{2}} \frac{- - \sin (x)}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3.4

Multiply $- 1$ by $- 1$ .

$lim_{x \to \frac{π}{2}} \frac{1 \sin (x)}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3.5

Multiply $\sin (x)$ by $1$ .

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{\frac{d}{d x} [- 2 \sin (2 x)]}$

Step 4.3.6

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

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 2 \frac{d}{d x} [\sin (2 x)]}$

Step 4.3.7

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

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Step 4.3.7.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 2 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])}$

Step 4.3.7.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 2 (\cos (u) \frac{d}{d x} [2 x])}$

Step 4.3.7.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 2 (\cos (2 x) \frac{d}{d x} [2 x])}$

Step 4.3.8

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

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 2 \cos (2 x) (2 \frac{d}{d x} [x])}$

Step 4.3.9

Multiply $2$ by $- 2$ .

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

Step 4.3.10

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

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

Step 4.3.11

Multiply $- 4$ by $1$ .

$lim_{x \to \frac{π}{2}} \frac{\sin (x)}{- 4 \cos (2 x)}$

Step 5

Evaluate the limit.

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Step 5.1

Move the term $\frac{1}{- 4}$ outside of the limit because it is constant with respect to $x$ .

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

Step 5.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $\frac{π}{2}$ .

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

Step 5.3

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

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

Step 5.4

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

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

Step 5.5

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

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

Step 6

Evaluate the limits by plugging in $\frac{π}{2}$ for all occurrences of $x$ .

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Step 6.1

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

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

Step 6.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

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

Step 7

Simplify the answer.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

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

Step 7.3

Simplify the denominator.

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Step 7.3.1

Cancel the common factor of $2$ .

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Step 7.3.1.1

Cancel the common factor.

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

Step 7.3.1.2

Rewrite the expression.

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

Step 7.3.2

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

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

Step 7.3.3

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

$- \frac{1}{4} \cdot \frac{1}{- 1 \cdot 1}$

Step 7.3.4

Multiply $- 1$ by $1$ .

$- \frac{1}{4} \cdot \frac{1}{- 1}$

Step 7.4

Move the negative one from the denominator of $\frac{1}{- 1}$ .

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

Step 7.5

Multiply $- 1$ by $1$ .

$- \frac{1}{4} \cdot - 1$

Step 7.6

Multiply $- \frac{1}{4} \cdot - 1$ .

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Step 7.6.1

Multiply $- 1$ by $- 1$ .

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

Step 7.6.2

Multiply $\frac{1}{4}$ by $1$ .

$\frac{1}{4}$