Calculus Examples

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

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.

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

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

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

Step 1.2.2

Move the limit into the exponent.

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

Step 1.2.3

Move the limit into the exponent.

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

Step 1.2.4

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

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

Step 1.2.5

Evaluate the limit of $2$ which is constant as $x$ approaches $0$ .

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

Step 1.2.6

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

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Step 1.2.6.1

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

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

Step 1.2.6.2

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

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

Step 1.2.7

Simplify the answer.

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Step 1.2.7.1

Simplify each term.

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Step 1.2.7.1.1

Anything raised to $0$ is $1$ .

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

Step 1.2.7.1.2

Anything raised to $0$ is $1$ .

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

Step 1.2.7.1.3

Multiply $- 1$ by $2$ .

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

Step 1.2.7.2

Add $1$ and $1$ .

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

Step 1.2.7.3

Subtract $2$ from $2$ .

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

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 $0$ .

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

Step 1.3.1.2

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

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

Step 1.3.1.3

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

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

Step 1.3.1.4

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

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

Step 1.3.2

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

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

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

Multiply $2$ by $0$ .

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

Step 1.3.3.1.2

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

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

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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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Step 3.4.1

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 $g (x) = - x$ .

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Step 3.4.1.1

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

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

Step 3.4.1.2

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

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

Step 3.4.1.3

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

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

Step 3.4.2

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

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

Step 3.4.3

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

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

Step 3.4.4

Multiply $- 1$ by $1$ .

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

Step 3.4.5

Move $- 1$ to the left of $e^{- x}$ .

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

Step 3.4.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

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

Step 3.5

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 3.6

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

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

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

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Step 3.9.1

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

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

Step 3.9.2

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

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Step 3.9.2.1

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

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

Step 3.9.2.2

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

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

Step 3.9.2.3

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

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

Step 3.9.3

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

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

Step 3.9.4

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

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

Step 3.9.5

Multiply $2$ by $1$ .

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

Step 3.9.6

Multiply $2$ by $- 1$ .

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

Step 3.9.7

Multiply $- 2$ by $- 1$ .

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

Step 3.10

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

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

Step 4

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

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

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

Step 5.1.2.2

Move the limit into the exponent.

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

Step 5.1.2.3

Move the limit into the exponent.

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

Step 5.1.2.4

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

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

Step 5.1.2.5

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

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Step 5.1.2.5.1

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

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

Step 5.1.2.5.2

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

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

Step 5.1.2.6

Simplify the answer.

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Step 5.1.2.6.1

Simplify each term.

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Step 5.1.2.6.1.1

Anything raised to $0$ is $1$ .

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

Step 5.1.2.6.1.2

Anything raised to $0$ is $1$ .

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

Step 5.1.2.6.1.3

Multiply $- 1$ by $1$ .

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

Step 5.1.2.6.2

Subtract $1$ from $1$ .

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

Step 5.1.3

Evaluate the limit of the denominator.

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Step 5.1.3.1

Evaluate the limit.

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Step 5.1.3.1.1

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

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

Step 5.1.3.1.2

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

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

Step 5.1.3.2

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

$\frac{1}{2} \cdot \frac{0}{\sin (2 \cdot 0)}$

Step 5.1.3.3

Simplify the answer.

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Step 5.1.3.3.1

Multiply $2$ by $0$ .

$\frac{1}{2} \cdot \frac{0}{\sin (0)}$

Step 5.1.3.3.2

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

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

Step 5.1.3.3.3

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

Undefined

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

Step 5.1.3.4

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

Undefined

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

Step 5.1.4

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

Undefined

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

Step 5.2

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.

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.3.4

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

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Step 5.3.4.1

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

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

Step 5.3.4.2

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 $g (x) = - x$ .

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Step 5.3.4.2.1

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

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

Step 5.3.4.2.2

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

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

Step 5.3.4.2.3

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

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

Step 5.3.4.3

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

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

Step 5.3.4.4

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

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

Step 5.3.4.5

Multiply $- 1$ by $1$ .

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

Step 5.3.4.6

Move $- 1$ to the left of $e^{- x}$ .

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

Step 5.3.4.7

Multiply $- 1$ by $- 1$ .

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

Step 5.3.4.8

Multiply $e^{- x}$ by $1$ .

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

Step 5.3.5

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 5.3.5.1

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

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

Step 5.3.5.2

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

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

Step 5.3.5.3

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

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

Step 5.3.6

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

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

Step 5.3.7

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

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

Step 5.3.8

Multiply $2$ by $1$ .

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

Step 5.3.9

Move $2$ to the left of $\cos (2 x)$ .

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

Step 6

Evaluate the limit.

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Step 6.1

Move the term $\frac{1}{2}$ outside of the limit because it is constant with respect to $x$ .

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

Step 6.2

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

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

Step 6.3

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

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

Step 6.4

Move the limit into the exponent.

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

Step 6.5

Move the limit into the exponent.

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

Step 6.6

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

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

Step 6.7

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

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

Step 6.8

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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

Step 7.3

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

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

Step 8

Simplify the answer.

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Step 8.1

Multiply $\frac{1}{2} \cdot \frac{1}{2}$ .

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Step 8.1.1

Multiply $\frac{1}{2}$ by $\frac{1}{2}$ .

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

Step 8.1.2

Multiply $2$ by $2$ .

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

Step 8.2

Simplify the numerator.

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Step 8.2.1

Anything raised to $0$ is $1$ .

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

Step 8.2.2

Anything raised to $0$ is $1$ .

$\frac{1}{4} \cdot \frac{1 + 1}{\cos (2 \cdot 0)}$

Step 8.2.3

Add $1$ and $1$ .

$\frac{1}{4} \cdot \frac{2}{\cos (2 \cdot 0)}$

Step 8.3

Simplify the denominator.

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Step 8.3.1

Multiply $2$ by $0$ .

$\frac{1}{4} \cdot \frac{2}{\cos (0)}$

Step 8.3.2

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

$\frac{1}{4} \cdot \frac{2}{1}$

Step 8.4

Cancel the common factor of $2$ .

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Step 8.4.1

Factor $2$ out of $4$ .

$\frac{1}{2 (2)} \cdot \frac{2}{1}$

Step 8.4.2

Cancel the common factor.

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

Step 8.4.3

Rewrite the expression.

$\frac{1}{2}$