Calculus Examples

$lim_{x \to \infty} {(\frac{3 x}{3 x + 1})}^{x}$

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(\frac{3 x}{3 x + 1})}^{x}$ as $e^{\ln ({(\frac{3 x}{3 x + 1})}^{x})}$ .

$lim_{x \to \infty} e^{\ln ({(\frac{3 x}{3 x + 1})}^{x})}$

Step 1.2

Expand $\ln ({(\frac{3 x}{3 x + 1})}^{x})$ by moving $x$ outside the logarithm.

$lim_{x \to \infty} e^{x \ln (\frac{3 x}{3 x + 1})}$

Step 2

Move the limit into the exponent.

$e^{lim_{x \to \infty} x \ln (\frac{3 x}{3 x + 1})}$

Step 3

Rewrite $x \ln (\frac{3 x}{3 x + 1})$ as $\frac{\ln (\frac{3 x}{3 x + 1})}{\frac{1}{x}}$ .

$e^{lim_{x \to \infty} \frac{\ln (\frac{3 x}{3 x + 1})}{\frac{1}{x}}}$

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.

$e^{\frac{lim_{x \to \infty} \ln (\frac{3 x}{3 x + 1})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2

Evaluate the limit of the numerator.

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Step 4.1.2.1

Evaluate the limit.

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Step 4.1.2.1.1

Move the limit inside the logarithm.

$e^{\frac{\ln (lim_{x \to \infty} \frac{3 x}{3 x + 1})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.1.2

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{x}{3 x + 1})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.2

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

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{3 x}{x} + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3

Evaluate the limit.

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Step 4.1.2.3.1

Cancel the common factor of $x$ .

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Step 4.1.2.3.1.1

Cancel the common factor.

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{3 x}{x} + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.1.2

Rewrite the expression.

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{1}{\frac{3 x}{x} + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.2

Cancel the common factor of $x$ .

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Step 4.1.2.3.2.1

Cancel the common factor.

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{1}{\frac{3 x}{x} + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.2.2

Divide $3$ by $1$ .

$e^{\frac{\ln (3 lim_{x \to \infty} \frac{1}{3 + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.3

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

$e^{\frac{\ln (3 \frac{lim_{x \to \infty} 1}{lim_{x \to \infty} 3 + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.4

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

$e^{\frac{\ln (3 \frac{1}{lim_{x \to \infty} 3 + \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.5

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

$e^{\frac{\ln (3 \frac{1}{lim_{x \to \infty} 3 + lim_{x \to \infty} \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.3.6

Evaluate the limit of $3$ which is constant as $x$ approaches $\infty$ .

$e^{\frac{\ln (3 \frac{1}{3 + lim_{x \to \infty} \frac{1}{x}})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.4

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

$e^{\frac{\ln (3 \frac{1}{3 + 0})}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.5

Simplify the answer.

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Step 4.1.2.5.1

Add $3$ and $0$ .

$e^{\frac{\ln (3 (\frac{1}{3}))}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.5.2

Cancel the common factor of $3$ .

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Step 4.1.2.5.2.1

Cancel the common factor.

$e^{\frac{\ln (3 (\frac{1}{3}))}{lim_{x \to \infty} \frac{1}{x}}}$

Step 4.1.2.5.2.2

Rewrite the expression.

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

Step 4.1.2.5.3

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

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

Step 4.1.3

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

$lim_{x \to \infty} \frac{\ln (\frac{3 x}{3 x + 1})}{\frac{1}{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (\frac{3 x}{3 x + 1})]}{\frac{d}{d x} [\frac{1}{x}]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$e^{lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (\frac{3 x}{3 x + 1})]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.2

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 $g (x) = \frac{3 x}{3 x + 1}$ .

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Step 4.3.2.1

To apply the Chain Rule, set $u$ as $\frac{3 x}{3 x + 1}$ .

$e^{lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [\frac{3 x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.2.2

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

$e^{lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [\frac{3 x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.2.3

Replace all occurrences of $u$ with $\frac{3 x}{3 x + 1}$ .

$e^{lim_{x \to \infty} \frac{\frac{1}{\frac{3 x}{3 x + 1}} \frac{d}{d x} [\frac{3 x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.3

Multiply by the reciprocal of the fraction to divide by $\frac{3 x}{3 x + 1}$ .

$e^{lim_{x \to \infty} \frac{1 \frac{3 x + 1}{3 x} \frac{d}{d x} [\frac{3 x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.4

Multiply $\frac{3 x + 1}{3 x}$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{3 x} \frac{d}{d x} [\frac{3 x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.5

Since $3$ is constant with respect to $x$ , the derivative of $\frac{3 x}{3 x + 1}$ with respect to $x$ is $3 \frac{d}{d x} [\frac{x}{3 x + 1}]$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{3 x} \cdot 3 \frac{d}{d x} [\frac{x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.6

Combine $3$ and $\frac{3 x + 1}{3 x}$ .

$e^{lim_{x \to \infty} \frac{\frac{3 (3 x + 1)}{3 x} \frac{d}{d x} [\frac{x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.7

Cancel the common factor of $3$ .

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Step 4.3.7.1

Cancel the common factor.

$e^{lim_{x \to \infty} \frac{\frac{3 (3 x + 1)}{3 x} \frac{d}{d x} [\frac{x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.7.2

Rewrite the expression.

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \frac{d}{d x} [\frac{x}{3 x + 1}]}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.8

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = 3 x + 1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{(3 x + 1) \frac{d}{d x} [x] - x \frac{d}{d x} [3 x + 1]}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.9

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

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{(3 x + 1) \cdot 1 - x \frac{d}{d x} [3 x + 1]}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.10

Multiply $3 x + 1$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x \frac{d}{d x} [3 x + 1]}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.11

By the Sum Rule, the derivative of $3 x + 1$ with respect to $x$ is $\frac{d}{d x} [3 x] + \frac{d}{d x} [1]$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x (\frac{d}{d x} [3 x] + \frac{d}{d x} [1])}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.12

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x (3 \frac{d}{d x} [x] + \frac{d}{d x} [1])}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.13

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

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x (3 \cdot 1 + \frac{d}{d x} [1])}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.14

Multiply $3$ by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x (3 + \frac{d}{d x} [1])}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.15

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

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x (3 + 0)}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.16

Add $3$ and $0$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - x \cdot 3}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.17

Multiply $3$ by $- 1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{3 x + 1 - 3 x}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.18

Subtract $3 x$ from $3 x$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{0 + 1}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.19

Add $0$ and $1$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x} \cdot \frac{1}{{(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.20

Multiply $\frac{3 x + 1}{x}$ by $\frac{1}{{(3 x + 1)}^{2}}$ .

$e^{lim_{x \to \infty} \frac{\frac{3 x + 1}{x {(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.21

Cancel the common factor of $3 x + 1$ and ${(3 x + 1)}^{2}$ .

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Step 4.3.21.1

Multiply by $1$ .

$e^{lim_{x \to \infty} \frac{\frac{(3 x + 1) \cdot 1}{x {(3 x + 1)}^{2}}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.21.2

Cancel the common factors.

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Step 4.3.21.2.1

Factor $3 x + 1$ out of $x {(3 x + 1)}^{2}$ .

$e^{lim_{x \to \infty} \frac{\frac{(3 x + 1) \cdot 1}{(3 x + 1) x (3 x + 1)}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.21.2.2

Cancel the common factor.

$e^{lim_{x \to \infty} \frac{\frac{(3 x + 1) \cdot 1}{(3 x + 1) x (3 x + 1)}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.21.2.3

Rewrite the expression.

$e^{lim_{x \to \infty} \frac{\frac{1}{x (3 x + 1)}}{\frac{d}{d x} [\frac{1}{x}]}}$

Step 4.3.22

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

$e^{lim_{x \to \infty} \frac{\frac{1}{x (3 x + 1)}}{\frac{d}{d x} [x^{- 1}]}}$

Step 4.3.23

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

$e^{lim_{x \to \infty} \frac{\frac{1}{x (3 x + 1)}}{- x^{- 2}}}$

Step 4.3.24

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

$e^{lim_{x \to \infty} \frac{\frac{1}{x (3 x + 1)}}{- \frac{1}{x^{2}}}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

$e^{lim_{x \to \infty} \frac{1}{x (3 x + 1)} \cdot - 1 x^{2}}$

Step 4.5

Combine $\frac{1}{x (3 x + 1)}$ and $x^{2}$ .

$e^{lim_{x \to \infty} - \frac{x^{2}}{x (3 x + 1)}}$

Step 4.6

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

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Step 4.6.1

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

$e^{lim_{x \to \infty} - \frac{x \cdot x}{x (3 x + 1)}}$

Step 4.6.2

Cancel the common factors.

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Step 4.6.2.1

Cancel the common factor.

$e^{lim_{x \to \infty} - \frac{x \cdot x}{x (3 x + 1)}}$

Step 4.6.2.2

Rewrite the expression.

$e^{lim_{x \to \infty} - \frac{x}{3 x + 1}}$

Step 5

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

$e^{- lim_{x \to \infty} \frac{x}{3 x + 1}}$

Step 6

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

$e^{- lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{3 x}{x} + \frac{1}{x}}}$

Step 7

Evaluate the limit.

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Step 7.1

Cancel the common factor of $x$ .

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Step 7.1.1

Cancel the common factor.

$e^{- lim_{x \to \infty} \frac{\frac{x}{x}}{\frac{3 x}{x} + \frac{1}{x}}}$

Step 7.1.2

Rewrite the expression.

$e^{- lim_{x \to \infty} \frac{1}{\frac{3 x}{x} + \frac{1}{x}}}$

Step 7.2

Cancel the common factor of $x$ .

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Step 7.2.1

Cancel the common factor.

$e^{- lim_{x \to \infty} \frac{1}{\frac{3 x}{x} + \frac{1}{x}}}$

Step 7.2.2

Divide $3$ by $1$ .

$e^{- lim_{x \to \infty} \frac{1}{3 + \frac{1}{x}}}$

Step 7.3

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

$e^{- \frac{lim_{x \to \infty} 1}{lim_{x \to \infty} 3 + \frac{1}{x}}}$

Step 7.4

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

$e^{- \frac{1}{lim_{x \to \infty} 3 + \frac{1}{x}}}$

Step 7.5

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

$e^{- \frac{1}{lim_{x \to \infty} 3 + lim_{x \to \infty} \frac{1}{x}}}$

Step 7.6

Evaluate the limit of $3$ which is constant as $x$ approaches $\infty$ .

$e^{- \frac{1}{3 + lim_{x \to \infty} \frac{1}{x}}}$

Step 8

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

$e^{- \frac{1}{3 + 0}}$

Step 9

Add $3$ and $0$ .

$e^{- \frac{1}{3}}$

Step 10

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

$\frac{1}{e^{\frac{1}{3}}}$