Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${\cos (2 x)}^{\frac{1}{x}}$ as $e^{\ln ({\cos (2 x)}^{\frac{1}{x}})}$ .

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

Step 1.2

Expand $\ln ({\cos (2 x)}^{\frac{1}{x}})$ by moving $\frac{1}{x}$ outside the logarithm.

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

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

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

Step 2.2

Combine $\frac{1}{x}$ and $\ln (\cos (2 x))$ .

$e^{lim_{x \to 0} \frac{\ln (\cos (2 x))}{x}}$

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.

$e^{\frac{lim_{x \to 0} \ln (\cos (2 x))}{lim_{x \to 0} x}}$

Step 3.1.2

Evaluate the limit of the numerator.

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Step 3.1.2.1

Evaluate the limit.

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Step 3.1.2.1.1

Move the limit inside the logarithm.

$e^{\frac{\ln (lim_{x \to 0} \cos (2 x))}{lim_{x \to 0} x}}$

Step 3.1.2.1.2

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

$e^{\frac{\ln (\cos (lim_{x \to 0} 2 x))}{lim_{x \to 0} x}}$

Step 3.1.2.1.3

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

$e^{\frac{\ln (\cos (2 lim_{x \to 0} x))}{lim_{x \to 0} x}}$

Step 3.1.2.2

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

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

Step 3.1.2.3

Simplify the answer.

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Step 3.1.2.3.1

Multiply $2$ by $0$ .

$e^{\frac{\ln (\cos (0))}{lim_{x \to 0} x}}$

Step 3.1.2.3.2

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

$e^{\frac{\ln (1)}{lim_{x \to 0} x}}$

Step 3.1.2.3.3

The natural logarithm of $1$ is $0$ .

$e^{\frac{0}{lim_{x \to 0} x}}$

Step 3.1.3

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

$e^{\frac{0}{0}}$

Step 3.1.4

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

Undefined

$e^{\frac{0}{0}}$

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.

$lim_{x \to 0} \frac{\ln (\cos (2 x))}{x} = lim_{x \to 0} \frac{\frac{d}{d x} [\ln (\cos (2 x))]}{\frac{d}{d x} [x]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

Step 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 $f (x) = \ln (x)$ and $g (x) = \cos (2 x)$ .

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Step 3.3.2.1

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

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

Step 3.3.2.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

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

Step 3.3.2.3

Replace all occurrences of $u_{1}$ with $\cos (2 x)$ .

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

Step 3.3.3

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 3.3.3.1

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

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

Step 3.3.3.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 3.3.3.3

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

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

Step 3.3.4

Combine $\frac{1}{\cos (2 x)}$ and $\sin (2 x)$ .

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

Step 3.3.5

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

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

Step 3.3.6

Multiply $2$ by $- 1$ .

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

Step 3.3.7

Combine $- 2$ and $\frac{\sin (2 x)}{\cos (2 x)}$ .

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

Step 3.3.8

Move the negative in front of the fraction.

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

Step 3.3.9

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

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

Step 3.3.10

Multiply $- 1$ by $1$ .

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

Step 3.3.11

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

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

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.5

Multiply $- 1$ by $1$ .

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

Step 4

Evaluate the limit.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

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

Step 4.5

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

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

Step 4.6

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

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

Step 4.7

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

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

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 $x$ .

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

Step 5.2

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

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

Step 6

Simplify the answer.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Multiply $2$ by $0$ .

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

Step 6.1.2

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

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

Step 6.2

Simplify the denominator.

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Step 6.2.1

Multiply $2$ by $0$ .

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

Step 6.2.2

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

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

Step 6.3

Divide $0$ by $1$ .

$e^{- 2 \cdot 0}$

Step 6.4

Multiply $- 2$ by $0$ .

$e^{0}$

Step 6.5

Anything raised to $0$ is $1$ .

$1$