Calculus Examples

lim x \to 0 {(1 + x)}^{\frac{1}{x}}

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

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite

${(1 + x)}^{\frac{1}{x}}$ as

$e^{\ln ({(1 + x)}^{\frac{1}{x}})}$ .

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

Step 1.2

Expand

$\ln ({(1 + x)}^{\frac{1}{x}})$ by moving

$\frac{1}{x}$ outside the logarithm.

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

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

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

Step 2.2

Combine

$\frac{1}{x}$ and

$\ln (1 + x)$ .

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

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.

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

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.

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

Step 3.1.2.1.2

Split the limit using the Sum of Limits Rule on the limit as

$x$ approaches

$0$ .

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

Step 3.1.2.1.3

Evaluate the limit of

$1$ which is constant as

$x$ approaches

$0$ .

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

Step 3.1.2.2

Evaluate the limit of

$x$ by plugging in

$0$ for

$x$ .

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

Step 3.1.2.3

Simplify the answer.

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Step 3.1.2.3.1

Add

$1$ and

$0$ .

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

Step 3.1.2.3.2

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.

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

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

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

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

$g (x) = 1 + x$ .

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Step 3.3.2.1

To apply the Chain Rule, set

$u$ as

$1 + x$ .

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

Step 3.3.2.2

The derivative of

$\ln (u)$ with respect to

$u$ is

$\frac{1}{u}$ .

$e^{lim_{x \to 0} \frac{\frac{1}{u} \frac{d}{d x} [1 + x]}{\frac{d}{d x} [x]}}$

Step 3.3.2.3

Replace all occurrences of

$u$ with

$1 + x$ .

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

Step 3.3.3

By the Sum Rule, the derivative of

$1 + x$ with respect to

$x$ is

$\frac{d}{d x} [1] + \frac{d}{d x} [x]$ .

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

Step 3.3.4

Since

$1$ is constant with respect to

$x$ , the derivative of

$1$ with respect to

$x$ is

$0$ .

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

Step 3.3.5

Add

$0$ and

$\frac{d}{d x} [x]$ .

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

Step 3.3.6

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 3.3.7

Multiply

$\frac{1}{1 + x}$ by

$1$ .

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

Step 3.3.8

Reorder terms.

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

Step 3.3.9

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.5

Multiply

$\frac{1}{x + 1}$ by

$1$ .

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

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

$0$ .

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

Step 4.2

Evaluate the limit of

$1$ which is constant as

$x$ approaches

$0$ .

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

Step 4.3

Split the limit using the Sum of Limits Rule on the limit as

$x$ approaches

$0$ .

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

Step 4.4

Evaluate the limit of

$1$ which is constant as

$x$ approaches

$0$ .

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

Step 5

Evaluate the limit of

$x$ by plugging in

$0$ for

$x$ .

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

Step 6

Simplify the answer.

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Step 6.1

Add

$0$ and

$1$ .

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

Step 6.2

Divide

$1$ by

$1$ .

$e^{1}$

Step 6.3

Simplify.

$e$

Step 7

The result can be shown in multiple forms.

Exact Form:

$e$

Decimal Form:

$2.71828182 \dots$