Calculus Examples

$lim_{x \to \infty} \frac{e^{2 x} + x^{2}}{e^{x} + 4 x}$

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.

$\frac{lim_{x \to \infty} e^{2 x} + x^{2}}{lim_{x \to \infty} e^{x} + 4 x}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

$\frac{lim_{x \to \infty} e^{2 x} + lim_{x \to \infty} x^{2}}{lim_{x \to \infty} e^{x} + 4 x}$

Step 1.2.2

Since the exponent $2 x$ approaches $\infty$ , the quantity $e^{2 x}$ approaches $\infty$ .

$\frac{\infty + lim_{x \to \infty} x^{2}}{lim_{x \to \infty} e^{x} + 4 x}$

Step 1.2.3

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

$\frac{\infty + \infty}{lim_{x \to \infty} e^{x} + 4 x}$

Step 1.2.4

Infinity plus infinity is infinity.

$\frac{\infty}{lim_{x \to \infty} e^{x} + 4 x}$

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

$\frac{\infty}{lim_{x \to \infty} e^{x} + lim_{x \to \infty} 4 x}$

Step 1.3.2

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

$\frac{\infty}{\infty + lim_{x \to \infty} 4 x}$

Step 1.3.3

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

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

Step 1.3.4

Infinity plus infinity is infinity.

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

Step 1.3.5

Infinity divided by infinity is undefined.

Undefined

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

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.

$lim_{x \to \infty} \frac{e^{2 x} + x^{2}}{e^{x} + 4 x} = lim_{x \to \infty} \frac{\frac{d}{d x} [e^{2 x} + x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [e^{2 x} + x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.2

By the Sum Rule, the derivative of $e^{2 x} + x^{2}$ with respect to $x$ is $\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [x^{2}]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 3.3.1

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 $g (x) = 2 x$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to \infty} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x] + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.1.2

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

$lim_{x \to \infty} \frac{e^{u} \frac{d}{d x} [2 x] + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.1.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to \infty} \frac{e^{2 x} \frac{d}{d x} [2 x] + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.3

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

$lim_{x \to \infty} \frac{e^{2 x} (2 \cdot 1) + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.4

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{e^{2 x} \cdot 2 + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.3.5

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to \infty} \frac{2 e^{2 x} + \frac{d}{d x} [x^{2}]}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to \infty} \frac{2 e^{2 x} + 2 x}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.5

Reorder terms.

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{\frac{d}{d x} [e^{x} + 4 x]}$

Step 3.6

By the Sum Rule, the derivative of $e^{x} + 4 x$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4 x]$ .

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4 x]}$

Step 3.7

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

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{e^{x} + \frac{d}{d x} [4 x]}$

Step 3.8

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 3.8.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{e^{x} + 4 \frac{d}{d x} [x]}$

Step 3.8.2

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

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{e^{x} + 4 \cdot 1}$

Step 3.8.3

Multiply $4$ by $1$ .

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{e^{x} + 4}$

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.

$\frac{lim_{x \to \infty} 2 x + 2 e^{2 x}}{lim_{x \to \infty} e^{x} + 4}$

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

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

$\frac{lim_{x \to \infty} 2 x + lim_{x \to \infty} 2 e^{2 x}}{lim_{x \to \infty} e^{x} + 4}$

Step 4.1.2.1.2

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

$\frac{\infty + lim_{x \to \infty} 2 e^{2 x}}{lim_{x \to \infty} e^{x} + 4}$

Step 4.1.2.2

Since the function $e^{2 x}$ approaches $\infty$ , the positive constant $2$ times the function also approaches $\infty$ .

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Step 4.1.2.2.1

Consider the limit with the constant multiple $2$ removed.

$\frac{\infty + lim_{x \to \infty} e^{2 x}}{lim_{x \to \infty} e^{x} + 4}$

Step 4.1.2.2.2

Since the exponent $2 x$ approaches $\infty$ , the quantity $e^{2 x}$ approaches $\infty$ .

$\frac{\infty + \infty}{lim_{x \to \infty} e^{x} + 4}$

Step 4.1.2.3

Infinity plus infinity is infinity.

$\frac{\infty}{lim_{x \to \infty} e^{x} + 4}$

Step 4.1.3

Evaluate the limit of the denominator.

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Step 4.1.3.1

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

$\frac{\infty}{lim_{x \to \infty} e^{x} + lim_{x \to \infty} 4}$

Step 4.1.3.2

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

$\frac{\infty}{\infty + lim_{x \to \infty} 4}$

Step 4.1.3.3

Evaluate the limit of $4$ which is constant as $x$ approaches $\infty$ .

$\frac{\infty}{\infty + 4}$

Step 4.1.3.4

Infinity plus or minus a number is infinity.

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

Step 4.1.3.5

Infinity divided by infinity is undefined.

Undefined

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

Step 4.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 4.2

$lim_{x \to \infty} \frac{2 x + 2 e^{2 x}}{e^{x} + 4} = lim_{x \to \infty} \frac{\frac{d}{d x} [2 x + 2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3

Find the derivative of the numerator and denominator.

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Step 4.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [2 x + 2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.2

By the Sum Rule, the derivative of $2 x + 2 e^{2 x}$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [2 e^{2 x}]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [2 x] + \frac{d}{d x} [2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.3

Evaluate $\frac{d}{d x} [2 x]$ .

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Step 4.3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{2 \frac{d}{d x} [x] + \frac{d}{d x} [2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.3.2

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

$lim_{x \to \infty} \frac{2 \cdot 1 + \frac{d}{d x} [2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.3.3

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{2 + \frac{d}{d x} [2 e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4

Evaluate $\frac{d}{d x} [2 e^{2 x}]$ .

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Step 4.3.4.1

Since $2$ is constant with respect to $x$ , the derivative of $2 e^{2 x}$ with respect to $x$ is $2 \frac{d}{d x} [e^{2 x}]$ .

$lim_{x \to \infty} \frac{2 + 2 \frac{d}{d x} [e^{2 x}]}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.2

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 $g (x) = 2 x$ .

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Step 4.3.4.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to \infty} \frac{2 + 2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.2.2

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

$lim_{x \to \infty} \frac{2 + 2 (e^{u} \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.2.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to \infty} \frac{2 + 2 (e^{2 x} \frac{d}{d x} [2 x])}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{2 + 2 (e^{2 x} (2 \frac{d}{d x} [x]))}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.4

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

$lim_{x \to \infty} \frac{2 + 2 (e^{2 x} (2 \cdot 1))}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.5

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{2 + 2 (e^{2 x} \cdot 2)}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.6

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to \infty} \frac{2 + 2 (2 \cdot e^{2 x})}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.4.7

Multiply $2$ by $2$ .

$lim_{x \to \infty} \frac{2 + 4 e^{2 x}}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.5

Reorder terms.

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{\frac{d}{d x} [e^{x} + 4]}$

Step 4.3.6

By the Sum Rule, the derivative of $e^{x} + 4$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4]$ .

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{\frac{d}{d x} [e^{x}] + \frac{d}{d x} [4]}$

Step 4.3.7

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

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{e^{x} + \frac{d}{d x} [4]}$

Step 4.3.8

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{e^{x} + 0}$

Step 4.3.9

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

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{e^{x}}$

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.

$\frac{lim_{x \to \infty} 4 e^{2 x} + 2}{lim_{x \to \infty} e^{x}}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

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

$\frac{lim_{x \to \infty} 4 e^{2 x} + lim_{x \to \infty} 2}{lim_{x \to \infty} e^{x}}$

Step 5.1.2.2

Since the function $e^{2 x}$ approaches $\infty$ , the positive constant $4$ times the function also approaches $\infty$ .

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Step 5.1.2.2.1

Consider the limit with the constant multiple $4$ removed.

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

Step 5.1.2.2.2

Since the exponent $2 x$ approaches $\infty$ , the quantity $e^{2 x}$ approaches $\infty$ .

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

Step 5.1.2.3

Evaluate the limit of $2$ which is constant as $x$ approaches $\infty$ .

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

Step 5.1.2.4

Infinity plus or minus a number is infinity.

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

Step 5.1.3

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

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

Step 5.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 5.2

$lim_{x \to \infty} \frac{4 e^{2 x} + 2}{e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [4 e^{2 x} + 2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [4 e^{2 x} + 2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.2

By the Sum Rule, the derivative of $4 e^{2 x} + 2$ with respect to $x$ is $\frac{d}{d x} [4 e^{2 x}] + \frac{d}{d x} [2]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [4 e^{2 x}] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3

Evaluate $\frac{d}{d x} [4 e^{2 x}]$ .

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Step 5.3.3.1

Since $4$ is constant with respect to $x$ , the derivative of $4 e^{2 x}$ with respect to $x$ is $4 \frac{d}{d x} [e^{2 x}]$ .

$lim_{x \to \infty} \frac{4 \frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.2

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 $g (x) = 2 x$ .

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Step 5.3.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$lim_{x \to \infty} \frac{4 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [2 x]) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.2.2

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

$lim_{x \to \infty} \frac{4 (e^{u} \frac{d}{d x} [2 x]) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.2.3

Replace all occurrences of $u$ with $2 x$ .

$lim_{x \to \infty} \frac{4 (e^{2 x} \frac{d}{d x} [2 x]) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.3

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{4 (e^{2 x} (2 \frac{d}{d x} [x])) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.4

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

$lim_{x \to \infty} \frac{4 (e^{2 x} (2 \cdot 1)) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.5

Multiply $2$ by $1$ .

$lim_{x \to \infty} \frac{4 (e^{2 x} \cdot 2) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.6

Move $2$ to the left of $e^{2 x}$ .

$lim_{x \to \infty} \frac{4 (2 \cdot e^{2 x}) + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.3.7

Multiply $2$ by $4$ .

$lim_{x \to \infty} \frac{8 e^{2 x} + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{x}]}$

Step 5.3.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{8 e^{2 x} + 0}{\frac{d}{d x} [e^{x}]}$

Step 5.3.5

Add $8 e^{2 x}$ and $0$ .

$lim_{x \to \infty} \frac{8 e^{2 x}}{\frac{d}{d x} [e^{x}]}$

Step 5.3.6

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

$lim_{x \to \infty} \frac{8 e^{2 x}}{e^{x}}$

Step 5.4

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

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Step 5.4.1

Factor $e^{x}$ out of $8 e^{2 x}$ .

$lim_{x \to \infty} \frac{e^{x} (8 e^{x})}{e^{x}}$

Step 5.4.2

Cancel the common factors.

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Step 5.4.2.1

Multiply by $1$ .

$lim_{x \to \infty} \frac{e^{x} (8 e^{x})}{e^{x} \cdot 1}$

Step 5.4.2.2

Cancel the common factor.

$lim_{x \to \infty} \frac{e^{x} (8 e^{x})}{e^{x} \cdot 1}$

Step 5.4.2.3

Rewrite the expression.

$lim_{x \to \infty} \frac{8 e^{x}}{1}$

Step 5.4.2.4

Divide $8 e^{x}$ by $1$ .

$lim_{x \to \infty} 8 e^{x}$

Step 6

Since the function $e^{x}$ approaches $\infty$ , the positive constant $8$ times the function also approaches $\infty$ .

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Step 6.1

Consider the limit with the constant multiple $8$ removed.

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

Step 6.2

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

$\infty$