Calculus Examples

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

Step 1

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

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Step 1.1

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

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

Step 1.2

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

$\frac{\infty}{lim_{x \to \infty} x}$

Step 1.3

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

$\frac{\infty}{\infty}$

Step 1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 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.

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

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

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

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

Step 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 3.4.1

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

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

Step 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 $a$ = $e$ .

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

Step 3.4.3

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

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

Step 3.5

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

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

Step 3.6

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

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

Step 3.7

Multiply $3$ by $1$ .

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

Simplify.

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Step 3.10.1

Reorder the factors of $\frac{1}{e^{3 x} + x} (3 e^{3 x} + 1)$ .

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

Step 3.10.2

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

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

Step 3.11

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

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

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

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

Step 6

Apply L'Hospital's rule.

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Step 6.1

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

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Step 6.1.1

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

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

Step 6.1.2

Evaluate the limit of the numerator.

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Step 6.1.2.1

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

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

Step 6.1.2.2

Since the function $e^{3 x}$ approaches $\infty$ , the positive constant $3$ times the function also approaches $\infty$ .

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Step 6.1.2.2.1

Consider the limit with the constant multiple $3$ removed.

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

Step 6.1.2.2.2

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

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

Step 6.1.2.3

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

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

Step 6.1.2.4

Infinity plus or minus a number is infinity.

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

Step 6.1.3

Evaluate the limit of the denominator.

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Step 6.1.3.1

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

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

Step 6.1.3.2

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

$\frac{\infty}{\infty + lim_{x \to \infty} x}$

Step 6.1.3.3

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

$\frac{\infty}{\infty + \infty}$

Step 6.1.3.4

Infinity plus infinity is infinity.

$\frac{\infty}{\infty}$

Step 6.1.3.5

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 6.1.4

Infinity divided by infinity is undefined.

Undefined

$\frac{\infty}{\infty}$

Step 6.2

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

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

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

Step 6.3.2

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

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

Step 6.3.3

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

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Step 6.3.3.1

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

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

Step 6.3.3.2

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

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Step 6.3.3.2.1

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

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

Step 6.3.3.2.2

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

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

Step 6.3.3.2.3

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

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

Step 6.3.3.3

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

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

Step 6.3.3.4

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

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

Step 6.3.3.5

Multiply $3$ by $1$ .

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

Step 6.3.3.6

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

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

Step 6.3.3.7

Multiply $3$ by $3$ .

$lim_{x \to \infty} \frac{9 e^{3 x} + \frac{d}{d x} [1]}{\frac{d}{d x} [e^{3 x} + x]}$

Step 6.3.4

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

$lim_{x \to \infty} \frac{9 e^{3 x} + 0}{\frac{d}{d x} [e^{3 x} + x]}$

Step 6.3.5

Add $9 e^{3 x}$ and $0$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{\frac{d}{d x} [e^{3 x} + x]}$

Step 6.3.6

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [x]}$

Step 6.3.7

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

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Step 6.3.7.1

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 6.3.7.1.1

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x] + \frac{d}{d x} [x]}$

Step 6.3.7.1.2

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{e^{u} \frac{d}{d x} [3 x] + \frac{d}{d x} [x]}$

Step 6.3.7.1.3

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [x]}$

Step 6.3.7.2

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{e^{3 x} (3 \frac{d}{d x} [x]) + \frac{d}{d x} [x]}$

Step 6.3.7.3

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{e^{3 x} (3 \cdot 1) + \frac{d}{d x} [x]}$

Step 6.3.7.4

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{e^{3 x} \cdot 3 + \frac{d}{d x} [x]}$

Step 6.3.7.5

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{3 e^{3 x} + \frac{d}{d x} [x]}$

Step 6.3.8

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{3 e^{3 x} + 1}$

Step 7

Move the term $9$ outside of the limit because it is constant with respect to $x$ .

$9 lim_{x \to \infty} \frac{e^{3 x}}{3 e^{3 x} + 1}$

Step 8

Apply L'Hospital's rule.

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Step 8.1

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

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Step 8.1.1

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

$9 \frac{lim_{x \to \infty} e^{3 x}}{lim_{x \to \infty} 3 e^{3 x} + 1}$

Step 8.1.2

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

$9 \frac{\infty}{lim_{x \to \infty} 3 e^{3 x} + 1}$

Step 8.1.3

Evaluate the limit of the denominator.

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Step 8.1.3.1

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

$9 \frac{\infty}{lim_{x \to \infty} 3 e^{3 x} + lim_{x \to \infty} 1}$

Step 8.1.3.2

Since the function $e^{3 x}$ approaches $\infty$ , the positive constant $3$ times the function also approaches $\infty$ .

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Step 8.1.3.2.1

Consider the limit with the constant multiple $3$ removed.

$9 \frac{\infty}{lim_{x \to \infty} e^{3 x} + lim_{x \to \infty} 1}$

Step 8.1.3.2.2

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

$9 \frac{\infty}{\infty + lim_{x \to \infty} 1}$

Step 8.1.3.3

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

$9 \frac{\infty}{\infty + 1}$

Step 8.1.3.4

Infinity plus or minus a number is infinity.

$9 \frac{\infty}{\infty}$

Step 8.1.3.5

Infinity divided by infinity is undefined.

Undefined

$9 \frac{\infty}{\infty}$

Step 8.1.4

Infinity divided by infinity is undefined.

Undefined

$9 \frac{\infty}{\infty}$

Step 8.2

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

Step 8.3

Find the derivative of the numerator and denominator.

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Step 8.3.1

Differentiate the numerator and denominator.

$9 lim_{x \to \infty} \frac{\frac{d}{d x} [e^{3 x}]}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.2

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 8.3.2.1

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

$9 lim_{x \to \infty} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x]}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.2.2

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

$9 lim_{x \to \infty} \frac{e^{u} \frac{d}{d x} [3 x]}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.2.3

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

$9 lim_{x \to \infty} \frac{e^{3 x} \frac{d}{d x} [3 x]}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.3

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

$9 lim_{x \to \infty} \frac{e^{3 x} \cdot 3 \frac{d}{d x} [x]}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.4

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

$9 lim_{x \to \infty} \frac{e^{3 x} \cdot 3 \cdot 1}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.5

Multiply $3$ by $1$ .

$9 lim_{x \to \infty} \frac{e^{3 x} \cdot 3}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.6

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

$9 lim_{x \to \infty} \frac{3 \cdot e^{3 x}}{\frac{d}{d x} [3 e^{3 x} + 1]}$

Step 8.3.7

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{\frac{d}{d x} [3 e^{3 x}] + \frac{d}{d x} [1]}$

Step 8.3.8

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

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Step 8.3.8.1

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 \frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [1]}$

Step 8.3.8.2

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 8.3.8.2.1

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 \frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x] + \frac{d}{d x} [1]}$

Step 8.3.8.2.2

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{u} \frac{d}{d x} [3 x] + \frac{d}{d x} [1]}$

Step 8.3.8.2.3

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [1]}$

Step 8.3.8.3

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} \cdot 3 \frac{d}{d x} [x] + \frac{d}{d x} [1]}$

Step 8.3.8.4

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} \cdot 3 \cdot 1 + \frac{d}{d x} [1]}$

Step 8.3.8.5

Multiply $3$ by $1$ .

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 e^{3 x} \cdot 3 + \frac{d}{d x} [1]}$

Step 8.3.8.6

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 \cdot 3 \cdot e^{3 x} + \frac{d}{d x} [1]}$

Step 8.3.8.7

Multiply $3$ by $3$ .

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{9 e^{3 x} + \frac{d}{d x} [1]}$

Step 8.3.9

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

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{9 e^{3 x} + 0}$

Step 8.3.10

Add $9 e^{3 x}$ and $0$ .

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{9 e^{3 x}}$

Step 8.4

Reduce.

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Step 8.4.1

Cancel the common factor of $3$ and $9$ .

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Step 8.4.1.1

Factor $3$ out of $3 e^{3 x}$ .

$9 lim_{x \to \infty} \frac{3 (e^{3 x})}{9 e^{3 x}}$

Step 8.4.1.2

Cancel the common factors.

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Step 8.4.1.2.1

Factor $3$ out of $9 e^{3 x}$ .

$9 lim_{x \to \infty} \frac{3 (e^{3 x})}{3 \cdot 3 e^{3 x}}$

Step 8.4.1.2.2

Cancel the common factor.

$9 lim_{x \to \infty} \frac{3 e^{3 x}}{3 \cdot 3 e^{3 x}}$

Step 8.4.1.2.3

Rewrite the expression.

$9 lim_{x \to \infty} \frac{e^{3 x}}{3 e^{3 x}}$

Step 8.4.2

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

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Step 8.4.2.1

Cancel the common factor.

$9 lim_{x \to \infty} \frac{e^{3 x}}{3 e^{3 x}}$

Step 8.4.2.2

Rewrite the expression.

$9 lim_{x \to \infty} \frac{1}{3}$

Step 9

Evaluate the limit.

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Step 9.1

Evaluate the limit of $\frac{1}{3}$ which is constant as $x$ approaches $\infty$ .

$9 (\frac{1}{3})$

Step 9.2

Cancel the common factor of $3$ .

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Step 9.2.1

Factor $3$ out of $9$ .

$3 (3) \frac{1}{3}$

Step 9.2.2

Cancel the common factor.

$3 \cdot 3 \frac{1}{3}$

Step 9.2.3

Rewrite the expression.

$3$