Calculus Examples

$lim_{x \to 0} {\tan (2 x)}^{x}$

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${\tan (2 x)}^{x}$ as $e^{\ln ({\tan (2 x)}^{x})}$ .

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

Step 1.2

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

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

Step 2

Set up the limit as a left-sided limit.

$lim_{x \to 0^{-}} e^{x \ln (\tan (2 x))}$

Step 3

Evaluate the limits by plugging in the value for the variable.

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Step 3.1

Evaluate the limit of $e^{x \ln (\tan (2 x))}$ by plugging in $0$ for $x$ .

$e^{0 \ln (\tan (2 \cdot 0))}$

Step 3.2

Multiply $2$ by $0$ .

$e^{0 \ln (\tan (0))}$

Step 3.3

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

$e^{0 \ln (0)}$

Step 3.4

Since $e^{0 \ln (0)}$ is undefined, the limit does not exist.

$Does not exist$

Step 4

Set up the limit as a right-sided limit.

$lim_{x \to 0^{+}} e^{x \ln (\tan (2 x))}$

Step 5

Evaluate the right-sided limit.

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Step 5.1

Move the limit into the exponent.

$e^{lim_{x \to 0^{+}} x \ln (\tan (2 x))}$

Step 5.2

Rewrite $x \ln (\tan (2 x))$ as $\frac{\ln (\tan (2 x))}{\frac{1}{x}}$ .

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

Step 5.3

Apply L'Hospital's rule.

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Step 5.3.1

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

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Step 5.3.1.1

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

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

Step 5.3.1.2

As $x$ approaches $0$ from the right side, $\ln (\tan (2 x))$ decreases without bound.

$e^{\frac{- \infty}{lim_{x \to 0^{+}} \frac{1}{x}}}$

Step 5.3.1.3

Since the numerator is a constant and the denominator approaches $0$ when $x$ approaches $0$ from the right, the fraction $\frac{1}{x}$ approaches infinity.

$e^{\frac{- \infty}{\infty}}$

Step 5.3.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{\frac{- \infty}{\infty}}$

Step 5.3.2

Since $\frac{- \infty}{\infty}$ 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.

$lim_{x \to 0^{+}} \frac{\ln (\tan (2 x))}{\frac{1}{x}} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} [\ln (\tan (2 x))]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 5.3.3

Find the derivative of the numerator and denominator.

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Step 5.3.3.1

Differentiate the numerator and denominator.

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

Step 5.3.3.2

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 $g (x) = \tan (2 x)$ .

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Step 5.3.3.2.1

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

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

Step 5.3.3.2.2

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

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

Step 5.3.3.2.3

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

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

Step 5.3.3.3

Rewrite $\tan (2 x)$ in terms of sines and cosines.

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

Step 5.3.3.4

Multiply by the reciprocal of the fraction to divide by $\frac{\sin (2 x)}{\cos (2 x)}$ .

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

Step 5.3.3.5

Write $1$ as a fraction with denominator $1$ .

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

Step 5.3.3.6

Simplify.

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Step 5.3.3.6.1

Rewrite the expression.

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

Step 5.3.3.6.2

Multiply $\frac{\cos (2 x)}{\sin (2 x)}$ by $1$ .

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

Step 5.3.3.7

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

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Step 5.3.3.7.1

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

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

Step 5.3.3.7.2

The derivative of $\tan (u_{2})$ with respect to $u_{2}$ is $\sec^{2} (u_{2})$ .

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

Step 5.3.3.7.3

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

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

Step 5.3.3.8

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

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

Step 5.3.3.9

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

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

Step 5.3.3.10

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

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

Step 5.3.3.11

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

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

Step 5.3.3.12

Multiply $\frac{2 \sec^{2} (2 x) \cos (2 x)}{\sin (2 x)}$ by $1$ .

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

Step 5.3.3.13

Simplify.

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Step 5.3.3.13.1

Simplify the numerator.

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Step 5.3.3.13.1.1

Rewrite $\sec (2 x)$ in terms of sines and cosines.

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

Step 5.3.3.13.1.2

Apply the product rule to $\frac{1}{\cos (2 x)}$ .

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

Step 5.3.3.13.1.3

One to any power is one.

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

Step 5.3.3.13.1.4

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

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

Step 5.3.3.13.1.5

Cancel the common factor of $\cos (2 x)$ .

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Step 5.3.3.13.1.5.1

Factor $\cos (2 x)$ out of $\cos^{2} (2 x)$ .

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

Step 5.3.3.13.1.5.2

Cancel the common factor.

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

Step 5.3.3.13.1.5.3

Rewrite the expression.

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

Step 5.3.3.13.2

Combine terms.

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Step 5.3.3.13.2.1

Rewrite $\frac{\frac{2}{\cos (2 x)}}{\sin (2 x)}$ as a product.

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

Step 5.3.3.13.2.2

Multiply $\frac{2}{\cos (2 x)}$ by $\frac{1}{\sin (2 x)}$ .

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

Step 5.3.3.14

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

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

Step 5.3.3.15

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

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

Step 5.3.3.16

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 5.3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.5

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

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

Step 5.4

Evaluate the limit.

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Step 5.4.1

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

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

Step 5.4.2

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

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

Step 5.5

Apply L'Hospital's rule.

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Step 5.5.1

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

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Step 5.5.1.1

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

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

Step 5.5.1.2

Evaluate the limit of the numerator.

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Step 5.5.1.2.1

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

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

Step 5.5.1.2.2

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

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

Step 5.5.1.2.3

Raising $0$ to any positive power yields $0$ .

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

Step 5.5.1.3

Evaluate the limit of the denominator.

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Step 5.5.1.3.1

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

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

Step 5.5.1.3.2

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

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

Step 5.5.1.3.3

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

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

Step 5.5.1.3.4

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

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

Step 5.5.1.3.5

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

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

Step 5.5.1.3.6

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

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Step 5.5.1.3.6.1

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

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

Step 5.5.1.3.6.2

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

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

Step 5.5.1.3.7

Simplify the answer.

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Step 5.5.1.3.7.1

Multiply $2$ by $0$ .

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

Step 5.5.1.3.7.2

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

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

Step 5.5.1.3.7.3

Multiply $\sin (2 \cdot 0)$ by $1$ .

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

Step 5.5.1.3.7.4

Multiply $2$ by $0$ .

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

Step 5.5.1.3.7.5

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

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

Step 5.5.1.3.7.6

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

Undefined

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

Step 5.5.1.3.8

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

Undefined

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

Step 5.5.1.4

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

Undefined

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

Step 5.5.2

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.

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

Step 5.5.3

Find the derivative of the numerator and denominator.

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Step 5.5.3.1

Differentiate the numerator and denominator.

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

Step 5.5.3.2

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

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

Step 5.5.3.3

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \cos (2 x)$ and $g (x) = \sin (2 x)$ .

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

Step 5.5.3.4

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 5.5.3.4.1

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

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

Step 5.5.3.4.2

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

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

Step 5.5.3.4.3

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

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

Step 5.5.3.5

Raise $\cos (2 x)$ to the power of $1$ .

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

Step 5.5.3.6

Raise $\cos (2 x)$ to the power of $1$ .

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

Step 5.5.3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.5.3.8

Add $1$ and $1$ .

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

Step 5.5.3.9

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

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

Step 5.5.3.10

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

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

Step 5.5.3.11

Multiply $2$ by $1$ .

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

Step 5.5.3.12

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

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

Step 5.5.3.13

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 5.5.3.13.1

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

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

Step 5.5.3.13.2

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

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

Step 5.5.3.13.3

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

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

Step 5.5.3.14

Raise $\sin (2 x)$ to the power of $1$ .

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

Step 5.5.3.15

Raise $\sin (2 x)$ to the power of $1$ .

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

Step 5.5.3.16

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.5.3.17

Add $1$ and $1$ .

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

Step 5.5.3.18

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

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

Step 5.5.3.19

Multiply $2$ by $- 1$ .

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

Step 5.5.3.20

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

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

Step 5.5.3.21

Multiply $- 2$ by $1$ .

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

Step 5.5.4

Cancel the common factor of $2$ and $2 \cos^{2} (2 x) - 2 \sin^{2} (2 x)$ .

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Step 5.5.4.1

Factor $2$ out of $2 x$ .

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

Step 5.5.4.2

Cancel the common factors.

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Step 5.5.4.2.1

Factor $2$ out of $2 \cos^{2} (2 x)$ .

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

Step 5.5.4.2.2

Factor $2$ out of $- 2 \sin^{2} (2 x)$ .

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

Step 5.5.4.2.3

Factor $2$ out of $2 (\cos^{2} (2 x)) + 2 (- \sin^{2} (2 x))$ .

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

Step 5.5.4.2.4

Cancel the common factor.

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

Step 5.5.4.2.5

Rewrite the expression.

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

Step 5.6

Evaluate the limit.

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Step 5.6.1

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

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

Step 5.6.2

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

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

Step 5.6.3

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

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

Step 5.6.4

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

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

Step 5.6.5

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

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

Step 5.6.6

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

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

Step 5.6.7

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

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

Step 5.6.8

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

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

Step 5.7

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

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Step 5.7.1

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

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

Step 5.7.2

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

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

Step 5.7.3

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

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

Step 5.8

Simplify the answer.

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Step 5.8.1

Simplify the denominator.

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Step 5.8.1.1

Multiply $2$ by $0$ .

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

Step 5.8.1.2

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

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

Step 5.8.1.3

One to any power is one.

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

Step 5.8.1.4

Multiply $2$ by $0$ .

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

Step 5.8.1.5

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

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

Step 5.8.1.6

Raising $0$ to any positive power yields $0$ .

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

Step 5.8.1.7

Multiply $- 1$ by $0$ .

$e^{- 2 \frac{0}{1 + 0}}$

Step 5.8.1.8

Add $1$ and $0$ .

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

Step 5.8.2

Divide $0$ by $1$ .

$e^{- 2 \cdot 0}$

Step 5.8.3

Multiply $- 2$ by $0$ .

$e^{0}$

Step 5.9

Anything raised to $0$ is $1$ .

$1$

Step 6

If either of the one-sided limits does not exist, the limit does not exist.

$Does not exist$