Calculus Examples

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

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

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

Step 1.2

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

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

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

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

Step 2.2

Combine $\frac{1}{x}$ and $\ln (2 - e^{2 x})$ .

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

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.

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

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.

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

Step 3.1.2.1.2

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

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

Step 3.1.2.1.3

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

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

Step 3.1.2.1.4

Move the limit into the exponent.

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

Step 3.1.2.1.5

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

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

Step 3.1.2.2

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

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

Step 3.1.2.3

Simplify the answer.

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Step 3.1.2.3.1

Simplify each term.

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Step 3.1.2.3.1.1

Multiply $2$ by $0$ .

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

Step 3.1.2.3.1.2

Anything raised to $0$ is $1$ .

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

Step 3.1.2.3.1.3

Multiply $- 1$ by $1$ .

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

Step 3.1.2.3.2

Subtract $1$ from $2$ .

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

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

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 $g (x) = 2 - e^{2 x}$ .

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Step 3.3.2.1

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

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

Step 3.3.2.2

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

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

Step 3.3.2.3

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

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

Step 3.3.3

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

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

Step 3.3.4

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

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

Step 3.3.5

Add $0$ and $\frac{d}{d x} [- e^{2 x}]$ .

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

Step 3.3.6

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

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

Step 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 $f (x) = e^{x}$ and $g (x) = 2 x$ .

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Step 3.3.7.1

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

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

Step 3.3.7.2

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

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

Step 3.3.7.3

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

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

Step 3.3.8

Combine $\frac{1}{2 - e^{2 x}}$ and $e^{2 x}$ .

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

Step 3.3.9

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

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

Step 3.3.10

Multiply $2$ by $- 1$ .

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

Step 3.3.11

Combine $- 2$ and $\frac{e^{2 x}}{2 - e^{2 x}}$ .

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

Step 3.3.12

Move the negative in front of the fraction.

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

Step 3.3.13

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

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

Step 3.3.14

Multiply $- 1$ by $1$ .

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

Step 3.3.15

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

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

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.5

Multiply $- 1$ by $1$ .

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

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 $x$ .

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

Move the limit into the exponent.

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

Step 4.5

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

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

Step 4.6

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

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

Step 4.7

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

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

Step 4.8

Move the limit into the exponent.

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

Step 4.9

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

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

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 $x$ .

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

Step 5.2

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

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

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 $0$ .

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

Step 6.1.2

Anything raised to $0$ is $1$ .

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

Step 6.2

Simplify the denominator.

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Step 6.2.1

Multiply $2$ by $0$ .

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

Step 6.2.2

Anything raised to $0$ is $1$ .

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

Step 6.2.3

Multiply $- 1$ by $1$ .

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

Step 6.2.4

Subtract $1$ from $2$ .

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

Step 6.3

Cancel the common factor of $1$ .

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Step 6.3.1

Cancel the common factor.

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

Step 6.3.2

Rewrite the expression.

$e^{- 2 \cdot 1}$

Step 6.4

Multiply $- 2$ by $1$ .

$e^{- 2}$

Step 6.5

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

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

Step 7

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.13533528 \dots$