Calculus Examples

$lim_{x \to 0} \frac{5 \sin (x)}{\cos^{2} (x) - 1}$

Step 1

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

Tap for more steps...

Step 1.1

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

$\frac{lim_{x \to 0} 5 \sin (x)}{lim_{x \to 0} \cos^{2} (x) - 1}$

Step 1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.2.1

Evaluate the limit.

Tap for more steps...

Step 1.2.1.1

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{5 lim_{x \to 0} \sin (x)}{lim_{x \to 0} \cos^{2} (x) - 1}$

Step 1.2.1.2

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

$\frac{5 \sin (lim_{x \to 0} x)}{lim_{x \to 0} \cos^{2} (x) - 1}$

Step 1.2.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

$\frac{5 \sin (0)}{lim_{x \to 0} \cos^{2} (x) - 1}$

Step 1.2.3

Simplify the answer.

Tap for more steps...

Step 1.2.3.1

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

$\frac{5 \cdot 0}{lim_{x \to 0} \cos^{2} (x) - 1}$

Step 1.2.3.2

Multiply $5$ by $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.3.1

Evaluate the limit.

Tap for more steps...

Step 1.3.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $0$ .

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

Step 1.3.1.2

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

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

Step 1.3.1.3

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

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

Step 1.3.1.4

Evaluate the limit of $1$ which is constant as $x$ approaches $0$ .

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

Step 1.3.2

Evaluate the limit of $x$ by plugging in $0$ for $x$ .

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

Step 1.3.3

Simplify the answer.

Tap for more steps...

Step 1.3.3.1

Simplify each term.

Tap for more steps...

Step 1.3.3.1.1

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

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

Step 1.3.3.1.2

One to any power is one.

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

Step 1.3.3.1.3

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.

$lim_{x \to 0} \frac{5 \sin (x)}{\cos^{2} (x) - 1} = lim_{x \to 0} \frac{\frac{d}{d x} [5 \sin (x)]}{\frac{d}{d x} [\cos^{2} (x) - 1]}$

Step 3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [5 \sin (x)]}{\frac{d}{d x} [\cos^{2} (x) - 1]}$

Step 3.2

Since $5$ is constant with respect to $x$ , the derivative of $5 \sin (x)$ with respect to $x$ is $5 \frac{d}{d x} [\sin (x)]$ .

$lim_{x \to 0} \frac{5 \frac{d}{d x} [\sin (x)]}{\frac{d}{d x} [\cos^{2} (x) - 1]}$

Step 3.3

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

$lim_{x \to 0} \frac{5 \cos (x)}{\frac{d}{d x} [\cos^{2} (x) - 1]}$

Step 3.4

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

$lim_{x \to 0} \frac{5 \cos (x)}{\frac{d}{d x} [\cos^{2} (x)] + \frac{d}{d x} [- 1]}$

Step 3.5

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

Tap for more steps...

Step 3.5.1

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

Tap for more steps...

Step 3.5.1.1

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

$lim_{x \to 0} \frac{5 \cos (x)}{\frac{d}{d u} [u^{2}] \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [- 1]}$

Step 3.5.1.2

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

$lim_{x \to 0} \frac{5 \cos (x)}{2 u \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [- 1]}$

Step 3.5.1.3

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

$lim_{x \to 0} \frac{5 \cos (x)}{2 \cos (x) \frac{d}{d x} [\cos (x)] + \frac{d}{d x} [- 1]}$

Step 3.5.2

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

$lim_{x \to 0} \frac{5 \cos (x)}{2 \cos (x) (- \sin (x)) + \frac{d}{d x} [- 1]}$

Step 3.5.3

Multiply $- 1$ by $2$ .

$lim_{x \to 0} \frac{5 \cos (x)}{- 2 \cos (x) \sin (x) + \frac{d}{d x} [- 1]}$

Step 3.6

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

$lim_{x \to 0} \frac{5 \cos (x)}{- 2 \cos (x) \sin (x) + 0}$

Step 3.7

Add $- 2 \cos (x) \sin (x)$ and $0$ .

$lim_{x \to 0} \frac{5 \cos (x)}{- 2 \cos (x) \sin (x)}$

Step 4

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

Tap for more steps...

Step 4.1

Cancel the common factor.

$lim_{x \to 0} \frac{5 \cos (x)}{- 2 \cos (x) \sin (x)}$

Step 4.2

Rewrite the expression.

$lim_{x \to 0} \frac{5}{- 2 \sin (x)}$

Step 5

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

$Does not exist$