Calculus Examples

$lim_{x \to \infty} {(e^{3 x} + 5)}^{\frac{2}{x}}$

Step 1

Use the properties of logarithms to simplify the limit.

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Step 1.1

Rewrite ${(e^{3 x} + 5)}^{\frac{2}{x}}$ as $e^{\ln ({(e^{3 x} + 5)}^{\frac{2}{x}})}$ .

$lim_{x \to \infty} e^{\ln ({(e^{3 x} + 5)}^{\frac{2}{x}})}$

Step 1.2

Expand $\ln ({(e^{3 x} + 5)}^{\frac{2}{x}})$ by moving $\frac{2}{x}$ outside the logarithm.

$lim_{x \to \infty} e^{\frac{2}{x} \ln (e^{3 x} + 5)}$

Step 2

Evaluate the limit.

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Step 2.1

Move the limit into the exponent.

$e^{lim_{x \to \infty} \frac{2}{x} \ln (e^{3 x} + 5)}$

Step 2.2

Combine $\frac{2}{x}$ and $\ln (e^{3 x} + 5)$ .

$e^{lim_{x \to \infty} \frac{2 \ln (e^{3 x} + 5)}{x}}$

Step 2.3

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

$e^{2 lim_{x \to \infty} \frac{\ln (e^{3 x} + 5)}{x}}$

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.

$e^{2 \frac{lim_{x \to \infty} \ln (e^{3 x} + 5)}{lim_{x \to \infty} x}}$

Step 3.1.2

As log approaches infinity, the value goes to $\infty$ .

$e^{2 \frac{\infty}{lim_{x \to \infty} x}}$

Step 3.1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

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

Step 3.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 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.

$lim_{x \to \infty} \frac{\ln (e^{3 x} + 5)}{x} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (e^{3 x} + 5)]}{\frac{d}{d x} [x]}$

Step 3.3

Find the derivative of the numerator and denominator.

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Step 3.3.1

Differentiate the numerator and denominator.

$e^{2 lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (e^{3 x} + 5)]}{\frac{d}{d x} [x]}}$

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 $g (x) = e^{3 x} + 5$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{1}$ as $e^{3 x} + 5$ .

$e^{2 lim_{x \to \infty} \frac{\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [e^{3 x} + 5]}{\frac{d}{d x} [x]}}$

Step 3.3.2.2

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

$e^{2 lim_{x \to \infty} \frac{\frac{1}{u_{1}} \frac{d}{d x} [e^{3 x} + 5]}{\frac{d}{d x} [x]}}$

Step 3.3.2.3

Replace all occurrences of $u_{1}$ with $e^{3 x} + 5$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} \frac{d}{d x} [e^{3 x} + 5]}{\frac{d}{d x} [x]}}$

Step 3.3.3

By the Sum Rule, the derivative of $e^{3 x} + 5$ with respect to $x$ is $\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [5]$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 3.3.4.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.4.2

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

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (e^{u_{2}} \frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.4.3

Replace all occurrences of $u_{2}$ with $3 x$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.5

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

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (e^{3 x} \cdot 3 \frac{d}{d x} [x] + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.6

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

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (e^{3 x} \cdot 3 \cdot 1 + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.7

Multiply $3$ by $1$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (e^{3 x} \cdot 3 + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.8

Move $3$ to the left of $e^{3 x}$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (3 \cdot e^{3 x} + \frac{d}{d x} [5])}{\frac{d}{d x} [x]}}$

Step 3.3.9

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} (3 e^{3 x} + 0)}{\frac{d}{d x} [x]}}$

Step 3.3.10

Add $3 e^{3 x}$ and $0$ .

$e^{2 lim_{x \to \infty} \frac{\frac{1}{e^{3 x} + 5} \cdot 3 e^{3 x}}{\frac{d}{d x} [x]}}$

Step 3.3.11

Combine $3$ and $\frac{1}{e^{3 x} + 5}$ .

$e^{2 lim_{x \to \infty} \frac{\frac{3}{e^{3 x} + 5} e^{3 x}}{\frac{d}{d x} [x]}}$

Step 3.3.12

Combine $\frac{3}{e^{3 x} + 5}$ and $e^{3 x}$ .

$e^{2 lim_{x \to \infty} \frac{\frac{3 e^{3 x}}{e^{3 x} + 5}}{\frac{d}{d x} [x]}}$

Step 3.3.13

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

$e^{2 lim_{x \to \infty} \frac{\frac{3 e^{3 x}}{e^{3 x} + 5}}{1}}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$e^{2 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} + 5} \cdot 1}$

Step 3.5

Multiply $\frac{3 e^{3 x}}{e^{3 x} + 5}$ by $1$ .

$e^{2 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} + 5}}$

Step 4

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x}}{e^{3 x} + 5}}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

$e^{2 \cdot 3 \frac{lim_{x \to \infty} e^{3 x}}{lim_{x \to \infty} e^{3 x} + 5}}$

Step 5.1.2

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

$e^{2 \cdot 3 \frac{\infty}{lim_{x \to \infty} e^{3 x} + 5}}$

Step 5.1.3

Evaluate the limit of the denominator.

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Step 5.1.3.1

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

$e^{2 \cdot 3 \frac{\infty}{lim_{x \to \infty} e^{3 x} + lim_{x \to \infty} 5}}$

Step 5.1.3.2

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

$e^{2 \cdot 3 \frac{\infty}{\infty + lim_{x \to \infty} 5}}$

Step 5.1.3.3

Evaluate the limit of $5$ which is constant as $x$ approaches $\infty$ .

$e^{2 \cdot 3 \frac{\infty}{\infty + 5}}$

Step 5.1.3.4

Infinity plus or minus a number is infinity.

$e^{2 \cdot 3 \frac{\infty}{\infty}}$

Step 5.1.3.5

Infinity divided by infinity is undefined.

Undefined

$e^{2 \cdot 3 \frac{\infty}{\infty}}$

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

$e^{2 \cdot 3 \frac{\infty}{\infty}}$

Step 5.2

$lim_{x \to \infty} \frac{e^{3 x}}{e^{3 x} + 5} = lim_{x \to \infty} \frac{\frac{d}{d x} [e^{3 x}]}{\frac{d}{d x} [e^{3 x} + 5]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$e^{2 \cdot 3 lim_{x \to \infty} \frac{\frac{d}{d x} [e^{3 x}]}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 5.3.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x]}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.2.2

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{u} \frac{d}{d x} [3 x]}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.2.3

Replace all occurrences of $u$ with $3 x$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x} \frac{d}{d x} [3 x]}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.3

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x} \cdot 3 \frac{d}{d x} [x]}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.4

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x} \cdot 3 \cdot 1}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.5

Multiply $3$ by $1$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x} \cdot 3}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.6

Move $3$ to the left of $e^{3 x}$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 \cdot e^{3 x}}{\frac{d}{d x} [e^{3 x} + 5]}}$

Step 5.3.7

By the Sum Rule, the derivative of $e^{3 x} + 5$ with respect to $x$ is $\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [5]$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [5]}}$

Step 5.3.8

Evaluate $\frac{d}{d x} [e^{3 x}]$ .

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Step 5.3.8.1

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 $g (x) = 3 x$ .

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Step 5.3.8.1.1

To apply the Chain Rule, set $u$ as $3 x$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x] + \frac{d}{d x} [5]}}$

Step 5.3.8.1.2

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{u} \frac{d}{d x} [3 x] + \frac{d}{d x} [5]}}$

Step 5.3.8.1.3

Replace all occurrences of $u$ with $3 x$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [5]}}$

Step 5.3.8.2

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} \cdot 3 \frac{d}{d x} [x] + \frac{d}{d x} [5]}}$

Step 5.3.8.3

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

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} \cdot 3 \cdot 1 + \frac{d}{d x} [5]}}$

Step 5.3.8.4

Multiply $3$ by $1$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{e^{3 x} \cdot 3 + \frac{d}{d x} [5]}}$

Step 5.3.8.5

Move $3$ to the left of $e^{3 x}$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} + \frac{d}{d x} [5]}}$

Step 5.3.9

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} + 0}}$

Step 5.3.10

Add $3 e^{3 x}$ and $0$ .

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x}}}$

Step 5.4

Reduce.

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Step 5.4.1

Cancel the common factor of $3$ .

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Step 5.4.1.1

Cancel the common factor.

$e^{2 \cdot 3 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x}}}$

Step 5.4.1.2

Rewrite the expression.

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x}}{e^{3 x}}}$

Step 5.4.2

Cancel the common factor of $e^{3 x}$ .

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Step 5.4.2.1

Cancel the common factor.

$e^{2 \cdot 3 lim_{x \to \infty} \frac{e^{3 x}}{e^{3 x}}}$

Step 5.4.2.2

Rewrite the expression.

$e^{2 \cdot 3 lim_{x \to \infty} 1}$

Step 6

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

$e^{2 \cdot 3 \cdot 1}$

Step 7

Multiply $2 \cdot 3 \cdot 1$ .

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Step 7.1

Multiply $2$ by $3$ .

$e^{6 \cdot 1}$

Step 7.2

Multiply $6$ by $1$ .

$e^{6}$