Calculus Examples

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

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.

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

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

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

Step 1.1.2.2

Move the limit into the exponent.

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

Step 1.1.2.3

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

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

Step 1.1.2.4

Move the limit into the exponent.

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

Step 1.1.2.5

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

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

Step 1.1.2.6

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.1.2.7

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.1.2.7.1

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

$\frac{e^{2 \cdot 0} - e^{- 2 lim_{x \to 0} x} - 4 lim_{x \to 0} x}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.7.2

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

$\frac{e^{2 \cdot 0} - e^{- 2 \cdot 0} - 4 lim_{x \to 0} x}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.7.3

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

$\frac{e^{2 \cdot 0} - e^{- 2 \cdot 0} - 4 \cdot 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.8

Simplify the answer.

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Step 1.1.2.8.1

Simplify each term.

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Step 1.1.2.8.1.1

Multiply $2$ by $0$ .

$\frac{e^{0} - e^{- 2 \cdot 0} - 4 \cdot 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.8.1.2

Anything raised to $0$ is $1$ .

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

Step 1.1.2.8.1.3

Multiply $- 2$ by $0$ .

$\frac{1 - e^{0} - 4 \cdot 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.8.1.4

Anything raised to $0$ is $1$ .

$\frac{1 - 1 \cdot 1 - 4 \cdot 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.8.1.5

Multiply $- 1$ by $1$ .

$\frac{1 - 1 - 4 \cdot 0}{lim_{x \to 0} x - \sin (x)}$

Step 1.1.2.8.1.6

Multiply $- 4$ by $0$ .

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

Step 1.1.2.8.2

Subtract $1$ from $1$ .

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

Step 1.1.2.8.3

Add $0$ and $0$ .

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

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 $0$ .

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 1.1.3.3.1

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

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

Step 1.1.3.3.2

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

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

Step 1.1.3.4

Simplify the answer.

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Step 1.1.3.4.1

Simplify each term.

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Step 1.1.3.4.1.1

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

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

Step 1.1.3.4.1.2

Multiply $- 1$ by $0$ .

$\frac{0}{0 + 0}$

Step 1.1.3.4.2

Add $0$ and $0$ .

$\frac{0}{0}$

Step 1.1.3.4.3

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

Undefined

$\frac{0}{0}$

Step 1.1.3.5

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.

$lim_{x \to 0} \frac{e^{2 x} - e^{- 2 x} - 4 x}{x - \sin (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [e^{2 x} - e^{- 2 x} - 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{2 x} - e^{- 2 x} - 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.2

By the Sum Rule, the derivative of $e^{2 x} - e^{- 2 x} - 4 x$ with respect to $x$ is $\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.3

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

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Step 1.3.3.1

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

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Step 1.3.3.1.1

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

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

Step 1.3.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 1.3.3.1.3

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

$lim_{x \to 0} \frac{e^{2 x} \frac{d}{d x} [2 x] + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.3.2

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

$lim_{x \to 0} \frac{e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.3.3

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

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

Step 1.3.3.4

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{e^{2 x} \cdot 2 + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.3.5

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to 0} \frac{2 e^{2 x} + \frac{d}{d x} [- e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4

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

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Step 1.3.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{- 2 x}$ with respect to $x$ is $- \frac{d}{d x} [e^{- 2 x}]$ .

$lim_{x \to 0} \frac{2 e^{2 x} - \frac{d}{d x} [e^{- 2 x}] + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.2

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

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Step 1.3.4.2.1

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

$lim_{x \to 0} \frac{2 e^{2 x} - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{2 e^{2 x} - (e^{u_{2}} \frac{d}{d x} [- 2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.2.3

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

$lim_{x \to 0} \frac{2 e^{2 x} - (e^{- 2 x} \frac{d}{d x} [- 2 x]) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.3

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

$lim_{x \to 0} \frac{2 e^{2 x} - (e^{- 2 x} (- 2 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.4

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

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

Step 1.3.4.5

Multiply $- 2$ by $1$ .

$lim_{x \to 0} \frac{2 e^{2 x} - (e^{- 2 x} \cdot - 2) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.6

Move $- 2$ to the left of $e^{- 2 x}$ .

$lim_{x \to 0} \frac{2 e^{2 x} - (- 2 \cdot e^{- 2 x}) + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.4.7

Multiply $- 2$ by $- 1$ .

$lim_{x \to 0} \frac{2 e^{2 x} + 2 e^{- 2 x} + \frac{d}{d x} [- 4 x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.5

Evaluate $\frac{d}{d x} [- 4 x]$ .

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Step 1.3.5.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$lim_{x \to 0} \frac{2 e^{2 x} + 2 e^{- 2 x} - 4 \frac{d}{d x} [x]}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.5.2

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

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

Step 1.3.5.3

Multiply $- 4$ by $1$ .

$lim_{x \to 0} \frac{2 e^{2 x} + 2 e^{- 2 x} - 4}{\frac{d}{d x} [x - \sin (x)]}$

Step 1.3.6

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

$lim_{x \to 0} \frac{2 e^{2 x} + 2 e^{- 2 x} - 4}{\frac{d}{d x} [x] + \frac{d}{d x} [- \sin (x)]}$

Step 1.3.7

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

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

Step 1.3.8

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

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Step 1.3.8.1

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

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

Step 1.3.8.2

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

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Evaluate the limit of the numerator.

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Step 2.1.2.1

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Move the limit into the exponent.

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

Step 2.1.2.4

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

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

Step 2.1.2.5

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

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

Step 2.1.2.6

Move the limit into the exponent.

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

Step 2.1.2.7

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

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

Step 2.1.2.8

Evaluate the limit of $4$ which is constant as $x$ approaches $0$ .

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

Step 2.1.2.9

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 2.1.2.9.1

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

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

Step 2.1.2.9.2

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

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

Step 2.1.2.10

Simplify the answer.

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Step 2.1.2.10.1

Simplify each term.

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Step 2.1.2.10.1.1

Multiply $2$ by $0$ .

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

Step 2.1.2.10.1.2

Anything raised to $0$ is $1$ .

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

Step 2.1.2.10.1.3

Multiply $2$ by $1$ .

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

Step 2.1.2.10.1.4

Multiply $- 2$ by $0$ .

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

Step 2.1.2.10.1.5

Anything raised to $0$ is $1$ .

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

Step 2.1.2.10.1.6

Multiply $2$ by $1$ .

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

Step 2.1.2.10.1.7

Multiply $- 1$ by $4$ .

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

Step 2.1.2.10.2

Add $2$ and $2$ .

$\frac{4 - 4}{lim_{x \to 0} 1 - \cos (x)}$

Step 2.1.2.10.3

Subtract $4$ from $4$ .

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

Step 2.1.3

Evaluate the limit of the denominator.

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Step 2.1.3.1

Evaluate the limit.

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Step 2.1.3.1.1

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

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

Step 2.1.3.1.2

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

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

Step 2.1.3.1.3

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

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

Step 2.1.3.2

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

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

Step 2.1.3.3

Simplify the answer.

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Step 2.1.3.3.1

Simplify each term.

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Step 2.1.3.3.1.1

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

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

Step 2.1.3.3.1.2

Multiply $- 1$ by $1$ .

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

Step 2.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 2.1.3.3.3

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

Undefined

$\frac{0}{0}$

Step 2.1.3.4

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

Undefined

$\frac{0}{0}$

Step 2.1.4

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

Undefined

$\frac{0}{0}$

Step 2.2

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.

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

Step 2.3.2

By the Sum Rule, the derivative of $2 e^{2 x} + 2 e^{- 2 x} - 4$ with respect to $x$ is $\frac{d}{d x} [2 e^{2 x}] + \frac{d}{d x} [2 e^{- 2 x}] + \frac{d}{d x} [- 4]$ .

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

Step 2.3.3

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

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Step 2.3.3.1

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

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

Step 2.3.3.2

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

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Step 2.3.3.2.1

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

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

Step 2.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3.3.2.3

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

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

Step 2.3.3.3

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

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

Step 2.3.3.4

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

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

Step 2.3.3.5

Multiply $2$ by $1$ .

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

Step 2.3.3.6

Move $2$ to the left of $e^{2 x}$ .

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

Step 2.3.3.7

Multiply $2$ by $2$ .

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

Step 2.3.4

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

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Step 2.3.4.1

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

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

Step 2.3.4.2

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

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Step 2.3.4.2.1

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

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

Step 2.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

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

Step 2.3.4.2.3

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

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

Step 2.3.4.3

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

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

Step 2.3.4.4

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

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

Step 2.3.4.5

Multiply $- 2$ by $1$ .

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

Step 2.3.4.6

Move $- 2$ to the left of $e^{- 2 x}$ .

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

Step 2.3.4.7

Multiply $- 2$ by $2$ .

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

Step 2.3.5

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

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

Step 2.3.6

Add $4 e^{2 x} - 4 e^{- 2 x}$ and $0$ .

$lim_{x \to 0} \frac{4 e^{2 x} - 4 e^{- 2 x}}{\frac{d}{d x} [1 - \cos (x)]}$

Step 2.3.7

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

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

Step 2.3.8

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

$lim_{x \to 0} \frac{4 e^{2 x} - 4 e^{- 2 x}}{0 + \frac{d}{d x} [- \cos (x)]}$

Step 2.3.9

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

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Step 2.3.9.1

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

$lim_{x \to 0} \frac{4 e^{2 x} - 4 e^{- 2 x}}{0 - \frac{d}{d x} [\cos (x)]}$

Step 2.3.9.2

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

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

Step 2.3.9.3

Multiply $- 1$ by $- 1$ .

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

Step 2.3.9.4

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

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

Step 2.3.10

Add $0$ and $\sin (x)$ .

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

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.

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

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

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

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

Step 3.1.2.2

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

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

Step 3.1.2.3

Move the limit into the exponent.

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

Step 3.1.2.4

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

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

Step 3.1.2.5

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

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

Step 3.1.2.6

Move the limit into the exponent.

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

Step 3.1.2.7

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

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

Step 3.1.2.8

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 3.1.2.8.1

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

$\frac{4 e^{2 \cdot 0} - 4 e^{- 2 lim_{x \to 0} x}}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.8.2

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

$\frac{4 e^{2 \cdot 0} - 4 e^{- 2 \cdot 0}}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9

Simplify the answer.

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Step 3.1.2.9.1

Simplify each term.

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Step 3.1.2.9.1.1

Multiply $2$ by $0$ .

$\frac{4 e^{0} - 4 e^{- 2 \cdot 0}}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9.1.2

Anything raised to $0$ is $1$ .

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

Step 3.1.2.9.1.3

Multiply $4$ by $1$ .

$\frac{4 - 4 e^{- 2 \cdot 0}}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9.1.4

Multiply $- 2$ by $0$ .

$\frac{4 - 4 e^{0}}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9.1.5

Anything raised to $0$ is $1$ .

$\frac{4 - 4 \cdot 1}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9.1.6

Multiply $- 4$ by $1$ .

$\frac{4 - 4}{lim_{x \to 0} \sin (x)}$

Step 3.1.2.9.2

Subtract $4$ from $4$ .

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

Step 3.1.3

Evaluate the limit of the denominator.

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Step 3.1.3.1

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

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

Step 3.1.3.2

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

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

Step 3.1.3.3

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

$\frac{0}{0}$

Step 3.1.3.4

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

Undefined

$\frac{0}{0}$

Step 3.1.4

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

Undefined

$\frac{0}{0}$

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.

$lim_{x \to 0} \frac{4 e^{2 x} - 4 e^{- 2 x}}{\sin (x)} = lim_{x \to 0} \frac{\frac{d}{d x} [4 e^{2 x} - 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$lim_{x \to 0} \frac{\frac{d}{d x} [4 e^{2 x} - 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.2

By the Sum Rule, the derivative of $4 e^{2 x} - 4 e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [4 e^{2 x}] + \frac{d}{d x} [- 4 e^{- 2 x}]$ .

$lim_{x \to 0} \frac{\frac{d}{d x} [4 e^{2 x}] + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3

Evaluate $\frac{d}{d x} [4 e^{2 x}]$ .

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Step 3.3.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 e^{2 x}$ with respect to $x$ is $4 \frac{d}{d x} [e^{2 x}]$ .

$lim_{x \to 0} \frac{4 \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3.2

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

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Step 3.3.3.2.1

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

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

Step 3.3.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 3.3.3.2.3

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

$lim_{x \to 0} \frac{4 (e^{2 x} \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3.3

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

$lim_{x \to 0} \frac{4 (e^{2 x} (2 \frac{d}{d x} [x])) + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3.4

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

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

Step 3.3.3.5

Multiply $2$ by $1$ .

$lim_{x \to 0} \frac{4 (e^{2 x} \cdot 2) + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3.6

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to 0} \frac{4 (2 \cdot e^{2 x}) + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.3.7

Multiply $2$ by $4$ .

$lim_{x \to 0} \frac{8 e^{2 x} + \frac{d}{d x} [- 4 e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4

Evaluate $\frac{d}{d x} [- 4 e^{- 2 x}]$ .

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Step 3.3.4.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 e^{- 2 x}$ with respect to $x$ is $- 4 \frac{d}{d x} [e^{- 2 x}]$ .

$lim_{x \to 0} \frac{8 e^{2 x} - 4 \frac{d}{d x} [e^{- 2 x}]}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.2

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

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Step 3.3.4.2.1

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

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (e^{u_{2}} \frac{d}{d x} [- 2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.2.3

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

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (e^{- 2 x} \frac{d}{d x} [- 2 x])}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.3

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

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (e^{- 2 x} (- 2 \frac{d}{d x} [x]))}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.4

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

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (e^{- 2 x} (- 2 \cdot 1))}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.5

Multiply $- 2$ by $1$ .

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (e^{- 2 x} \cdot - 2)}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.6

Move $- 2$ to the left of $e^{- 2 x}$ .

$lim_{x \to 0} \frac{8 e^{2 x} - 4 (- 2 \cdot e^{- 2 x})}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.4.7

Multiply $- 2$ by $- 4$ .

$lim_{x \to 0} \frac{8 e^{2 x} + 8 e^{- 2 x}}{\frac{d}{d x} [\sin (x)]}$

Step 3.3.5

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

$lim_{x \to 0} \frac{8 e^{2 x} + 8 e^{- 2 x}}{\cos (x)}$

Step 4

Evaluate the limit.

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Step 4.1

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

$\frac{lim_{x \to 0} 8 e^{2 x} + 8 e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.2

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

$\frac{lim_{x \to 0} 8 e^{2 x} + lim_{x \to 0} 8 e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.3

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{8 lim_{x \to 0} e^{2 x} + lim_{x \to 0} 8 e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.4

Move the limit into the exponent.

$\frac{8 e^{lim_{x \to 0} 2 x} + lim_{x \to 0} 8 e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.5

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

$\frac{8 e^{2 lim_{x \to 0} x} + lim_{x \to 0} 8 e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.6

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{8 e^{2 lim_{x \to 0} x} + 8 lim_{x \to 0} e^{- 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.7

Move the limit into the exponent.

$\frac{8 e^{2 lim_{x \to 0} x} + 8 e^{lim_{x \to 0} - 2 x}}{lim_{x \to 0} \cos (x)}$

Step 4.8

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

$\frac{8 e^{2 lim_{x \to 0} x} + 8 e^{- 2 lim_{x \to 0} x}}{lim_{x \to 0} \cos (x)}$

Step 4.9

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

$\frac{8 e^{2 lim_{x \to 0} x} + 8 e^{- 2 lim_{x \to 0} x}}{\cos (lim_{x \to 0} x)}$

Step 5

Evaluate the limits by plugging in $0$ for all occurrences of $x$ .

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Step 5.1

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

$\frac{8 e^{2 \cdot 0} + 8 e^{- 2 lim_{x \to 0} x}}{\cos (lim_{x \to 0} x)}$

Step 5.2

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

$\frac{8 e^{2 \cdot 0} + 8 e^{- 2 \cdot 0}}{\cos (lim_{x \to 0} x)}$

Step 5.3

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

$\frac{8 e^{2 \cdot 0} + 8 e^{- 2 \cdot 0}}{\cos (0)}$

Step 6

Simplify the answer.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Factor $8$ out of $8 e^{2 \cdot 0} + 8 e^{- 2 \cdot 0}$ .

$\frac{8 (e^{2 \cdot 0} + e^{- 2 \cdot 0})}{\cos (0)}$

Step 6.1.2

Multiply $2$ by $0$ .

$\frac{8 (e^{0} + e^{- 2 \cdot 0})}{\cos (0)}$

Step 6.1.3

Anything raised to $0$ is $1$ .

$\frac{8 (1 + e^{- 2 \cdot 0})}{\cos (0)}$

Step 6.1.4

Multiply $- 2$ by $0$ .

$\frac{8 (1 + e^{0})}{\cos (0)}$

Step 6.1.5

Anything raised to $0$ is $1$ .

$\frac{8 (1 + 1)}{\cos (0)}$

Step 6.1.6

Add $1$ and $1$ .

$\frac{8 \cdot 2}{\cos (0)}$

Step 6.2

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

$\frac{8 \cdot 2}{1}$

Step 6.3

Multiply $8$ by $2$ .

$\frac{16}{1}$

Step 6.4

Divide $16$ by $1$ .

$16$