Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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

Step 1.1.2.3

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

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

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 $\frac{π}{2}$ .

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

Step 1.1.3.1.2

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

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

$\frac{0}{1 - \sin (\frac{π}{2})}$

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

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

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

Step 1.1.3.3.1.2

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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

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

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Step 1.3.5.1

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

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

Step 1.3.5.2

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

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

Step 1.3.6

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

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

Step 1.4

Dividing two negative values results in a positive value.

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

Step 2

Since the function approaches $\infty$ from the left and $- \infty$ from the right, the limit does not exist.

$Does not exist$