Calculus Examples

$lim_{n \to \infty} {(\frac{n}{n + 1})}^{n}$

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(\frac{n}{n + 1})}^{n}$ as $e^{\ln ({(\frac{n}{n + 1})}^{n})}$ .

$lim_{n \to \infty} e^{\ln ({(\frac{n}{n + 1})}^{n})}$

Step 1.2

Expand $\ln ({(\frac{n}{n + 1})}^{n})$ by moving $n$ outside the logarithm.

$lim_{n \to \infty} e^{n \ln (\frac{n}{n + 1})}$

Step 2

Move the limit into the exponent.

$e^{lim_{n \to \infty} n \ln (\frac{n}{n + 1})}$

Step 3

Rewrite $n \ln (\frac{n}{n + 1})$ as $\frac{\ln (\frac{n}{n + 1})}{\frac{1}{n}}$ .

$e^{lim_{n \to \infty} \frac{\ln (\frac{n}{n + 1})}{\frac{1}{n}}}$

Step 4

Apply L'Hospital's rule.

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Step 4.1

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

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Step 4.1.1

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

$e^{\frac{lim_{n \to \infty} \ln (\frac{n}{n + 1})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Move the limit inside the logarithm.

$e^{\frac{\ln (lim_{n \to \infty} \frac{n}{n + 1})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.2

Divide the numerator and denominator by the highest power of $n$ in the denominator, which is $n$ .

$e^{\frac{\ln (lim_{n \to \infty} \frac{\frac{n}{n}}{\frac{n}{n} + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3

Evaluate the limit.

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Step 4.1.2.3.1

Cancel the common factor of $n$ .

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Step 4.1.2.3.1.1

Cancel the common factor.

$e^{\frac{\ln (lim_{n \to \infty} \frac{\frac{n}{n}}{\frac{n}{n} + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.1.2

Rewrite the expression.

$e^{\frac{\ln (lim_{n \to \infty} \frac{1}{\frac{n}{n} + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.2

Cancel the common factor of $n$ .

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Step 4.1.2.3.2.1

Cancel the common factor.

$e^{\frac{\ln (lim_{n \to \infty} \frac{1}{\frac{n}{n} + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.2.2

Rewrite the expression.

$e^{\frac{\ln (lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.3

Split the limit using the Limits Quotient Rule on the limit as $n$ approaches $\infty$ .

$e^{\frac{\ln (\frac{lim_{n \to \infty} 1}{lim_{n \to \infty} 1 + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.4

Evaluate the limit of $1$ which is constant as $n$ approaches $\infty$ .

$e^{\frac{\ln (\frac{1}{lim_{n \to \infty} 1 + \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.5

Split the limit using the Sum of Limits Rule on the limit as $n$ approaches $\infty$ .

$e^{\frac{\ln (\frac{1}{lim_{n \to \infty} 1 + lim_{n \to \infty} \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.3.6

Evaluate the limit of $1$ which is constant as $n$ approaches $\infty$ .

$e^{\frac{\ln (\frac{1}{1 + lim_{n \to \infty} \frac{1}{n}})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.4

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{n}$ approaches $0$ .

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

Step 4.1.2.5

Simplify the answer.

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Step 4.1.2.5.1

Add $1$ and $0$ .

$e^{\frac{\ln (\frac{1}{1})}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.5.2

Divide $1$ by $1$ .

$e^{\frac{\ln (1)}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.2.5.3

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

$e^{\frac{0}{lim_{n \to \infty} \frac{1}{n}}}$

Step 4.1.3

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{n}$ approaches $0$ .

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

Step 4.1.4

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

Undefined

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

Step 4.2

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.

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

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

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

Step 4.3.2

Differentiate using the chain rule, which states that $\frac{d}{d n} [f (g (n))]$ is $f' (g (n)) g' (n)$ where $f (n) = \ln (n)$ and $g (n) = \frac{n}{n + 1}$ .

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Step 4.3.2.1

To apply the Chain Rule, set $u$ as $\frac{n}{n + 1}$ .

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

Step 4.3.2.2

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

$e^{lim_{n \to \infty} \frac{\frac{1}{u} \frac{d}{d n} [\frac{n}{n + 1}]}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.2.3

Replace all occurrences of $u$ with $\frac{n}{n + 1}$ .

$e^{lim_{n \to \infty} \frac{\frac{1}{\frac{n}{n + 1}} \frac{d}{d n} [\frac{n}{n + 1}]}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{n}{n + 1}$ .

$e^{lim_{n \to \infty} \frac{1 \frac{n + 1}{n} \frac{d}{d n} [\frac{n}{n + 1}]}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.4

Multiply $\frac{n + 1}{n}$ by $1$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \frac{d}{d n} [\frac{n}{n + 1}]}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.5

Differentiate using the Quotient Rule which states that $\frac{d}{d n} [\frac{f (n)}{g (n)}]$ is $\frac{g (n) \frac{d}{d n} [f (n)] - f (n) \frac{d}{d n} [g (n)]}{{g (n)}^{2}}$ where $f (n) = n$ and $g (n) = n + 1$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{(n + 1) \frac{d}{d n} [n] - n \frac{d}{d n} [n + 1]}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.6

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

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{(n + 1) \cdot 1 - n \frac{d}{d n} [n + 1]}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.7

Multiply $n + 1$ by $1$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{n + 1 - n \frac{d}{d n} [n + 1]}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.8

By the Sum Rule, the derivative of $n + 1$ with respect to $n$ is $\frac{d}{d n} [n] + \frac{d}{d n} [1]$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{n + 1 - n (\frac{d}{d n} [n] + \frac{d}{d n} [1])}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.9

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

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{n + 1 - n (1 + \frac{d}{d n} [1])}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.10

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

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

Step 4.3.11

Add $1$ and $0$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{n + 1 - n \cdot 1}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.12

Multiply $- 1$ by $1$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{n + 1 - n}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.13

Subtract $n$ from $n$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{0 + 1}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.14

Add $0$ and $1$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n} \cdot \frac{1}{{(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.15

Multiply $\frac{n + 1}{n}$ by $\frac{1}{{(n + 1)}^{2}}$ .

$e^{lim_{n \to \infty} \frac{\frac{n + 1}{n {(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.16

Cancel the common factor of $n + 1$ and ${(n + 1)}^{2}$ .

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Step 4.3.16.1

Multiply by $1$ .

$e^{lim_{n \to \infty} \frac{\frac{(n + 1) \cdot 1}{n {(n + 1)}^{2}}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.16.2

Cancel the common factors.

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Step 4.3.16.2.1

Factor $n + 1$ out of $n {(n + 1)}^{2}$ .

$e^{lim_{n \to \infty} \frac{\frac{(n + 1) \cdot 1}{(n + 1) n (n + 1)}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.16.2.2

Cancel the common factor.

$e^{lim_{n \to \infty} \frac{\frac{(n + 1) \cdot 1}{(n + 1) n (n + 1)}}{\frac{d}{d n} [\frac{1}{n}]}}$

Step 4.3.16.2.3

Rewrite the expression.

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

Step 4.3.17

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

$e^{lim_{n \to \infty} \frac{\frac{1}{n (n + 1)}}{\frac{d}{d n} [n^{- 1}]}}$

Step 4.3.18

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

$e^{lim_{n \to \infty} \frac{\frac{1}{n (n + 1)}}{- n^{- 2}}}$

Step 4.3.19

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

$e^{lim_{n \to \infty} \frac{\frac{1}{n (n + 1)}}{- \frac{1}{n^{2}}}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

$e^{lim_{n \to \infty} \frac{1}{n (n + 1)} \cdot - 1 n^{2}}$

Step 4.5

Combine $\frac{1}{n (n + 1)}$ and $n^{2}$ .

$e^{lim_{n \to \infty} - \frac{n^{2}}{n (n + 1)}}$

Step 4.6

Cancel the common factor of $n^{2}$ and $n$ .

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Step 4.6.1

Factor $n$ out of $n^{2}$ .

$e^{lim_{n \to \infty} - \frac{n \cdot n}{n (n + 1)}}$

Step 4.6.2

Cancel the common factors.

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Step 4.6.2.1

Cancel the common factor.

$e^{lim_{n \to \infty} - \frac{n \cdot n}{n (n + 1)}}$

Step 4.6.2.2

Rewrite the expression.

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

Step 5

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

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

Step 6

Divide the numerator and denominator by the highest power of $n$ in the denominator, which is $n$ .

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

Step 7

Evaluate the limit.

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Step 7.1

Cancel the common factor of $n$ .

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Step 7.1.1

Cancel the common factor.

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

Step 7.1.2

Rewrite the expression.

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

Step 7.2

Cancel the common factor of $n$ .

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Step 7.2.1

Cancel the common factor.

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

Step 7.2.2

Rewrite the expression.

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

Step 7.3

Split the limit using the Limits Quotient Rule on the limit as $n$ approaches $\infty$ .

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

Step 7.4

Evaluate the limit of $1$ which is constant as $n$ approaches $\infty$ .

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

Step 7.5

Split the limit using the Sum of Limits Rule on the limit as $n$ approaches $\infty$ .

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

Step 7.6

Evaluate the limit of $1$ which is constant as $n$ approaches $\infty$ .

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

Step 8

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{n}$ approaches $0$ .

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

Step 9

Simplify the answer.

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Step 9.1

Add $1$ and $0$ .

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

Step 9.2

Cancel the common factor of $1$ .

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Step 9.2.1

Cancel the common factor.

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

Step 9.2.2

Rewrite the expression.

$e^{- 1 \cdot 1}$

Step 9.3

Multiply $- 1$ by $1$ .

$e^{- 1}$

Step 10

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

$\frac{1}{e}$

Step 11

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{e}$

Decimal Form:

$0.36787944 \dots$