Calculus Examples

$lim_{x \to \infty} \frac{4 x^{2} - 5 x + 2}{e^{5 x} + \ln (x)}$

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.

$\frac{lim_{x \to \infty} 4 x^{2} - 5 x + 2}{lim_{x \to \infty} e^{5 x} + \ln (x)}$

Step 1.2

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

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

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 $\infty$ .

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

Step 1.3.2

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

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

Step 1.3.3

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

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

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.

$lim_{x \to \infty} \frac{4 x^{2} - 5 x + 2}{e^{5 x} + \ln (x)} = lim_{x \to \infty} \frac{\frac{d}{d x} [4 x^{2} - 5 x + 2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [4 x^{2} - 5 x + 2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.2

By the Sum Rule, the derivative of $4 x^{2} - 5 x + 2$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [2]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.3

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

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Step 3.3.1

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

$lim_{x \to \infty} \frac{4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.3.2

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

$lim_{x \to \infty} \frac{4 (2 x) + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.3.3

Multiply $2$ by $4$ .

$lim_{x \to \infty} \frac{8 x + \frac{d}{d x} [- 5 x] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.4

Evaluate $\frac{d}{d x} [- 5 x]$ .

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Step 3.4.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{8 x - 5 \frac{d}{d x} [x] + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.4.2

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

$lim_{x \to \infty} \frac{8 x - 5 \cdot 1 + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.4.3

Multiply $- 5$ by $1$ .

$lim_{x \to \infty} \frac{8 x - 5 + \frac{d}{d x} [2]}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.5

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

$lim_{x \to \infty} \frac{8 x - 5 + 0}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.6

Add $8 x - 5$ and $0$ .

$lim_{x \to \infty} \frac{8 x - 5}{\frac{d}{d x} [e^{5 x} + \ln (x)]}$

Step 3.7

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

$lim_{x \to \infty} \frac{8 x - 5}{\frac{d}{d x} [e^{5 x}] + \frac{d}{d x} [\ln (x)]}$

Step 3.8

Evaluate $\frac{d}{d x} [e^{5 x}]$ .

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Step 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 $g (x) = 5 x$ .

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Step 3.8.1.1

To apply the Chain Rule, set $u$ as $5 x$ .

$lim_{x \to \infty} \frac{8 x - 5}{\frac{d}{d u} [e^{u}] \frac{d}{d x} [5 x] + \frac{d}{d x} [\ln (x)]}$

Step 3.8.1.2

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

$lim_{x \to \infty} \frac{8 x - 5}{e^{u} \frac{d}{d x} [5 x] + \frac{d}{d x} [\ln (x)]}$

Step 3.8.1.3

Replace all occurrences of $u$ with $5 x$ .

$lim_{x \to \infty} \frac{8 x - 5}{e^{5 x} \frac{d}{d x} [5 x] + \frac{d}{d x} [\ln (x)]}$

Step 3.8.2

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{8 x - 5}{e^{5 x} (5 \frac{d}{d x} [x]) + \frac{d}{d x} [\ln (x)]}$

Step 3.8.3

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

$lim_{x \to \infty} \frac{8 x - 5}{e^{5 x} (5 \cdot 1) + \frac{d}{d x} [\ln (x)]}$

Step 3.8.4

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{8 x - 5}{e^{5 x} \cdot 5 + \frac{d}{d x} [\ln (x)]}$

Step 3.8.5

Move $5$ to the left of $e^{5 x}$ .

$lim_{x \to \infty} \frac{8 x - 5}{5 e^{5 x} + \frac{d}{d x} [\ln (x)]}$

Step 3.9

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

$lim_{x \to \infty} \frac{8 x - 5}{5 e^{5 x} + \frac{1}{x}}$

Step 3.10

Reorder terms.

$lim_{x \to \infty} \frac{8 x - 5}{\frac{1}{x} + 5 e^{5 x}}$

Step 4

Combine terms.

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Step 4.1

To write $5 e^{5 x}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$lim_{x \to \infty} \frac{8 x - 5}{\frac{1}{x} + \frac{5 e^{5 x} x}{x}}$

Step 4.2

Combine the numerators over the common denominator.

$lim_{x \to \infty} \frac{8 x - 5}{\frac{1 + 5 e^{5 x} x}{x}}$

Step 5

Simplify the limit argument.

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Step 5.1

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} (8 x - 5) \frac{x}{1 + 5 e^{5 x} x}$

Step 5.2

Multiply $8 x - 5$ by $\frac{x}{1 + 5 e^{5 x} x}$ .

$lim_{x \to \infty} \frac{(8 x - 5) x}{1 + 5 e^{5 x} x}$

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.

$\frac{lim_{x \to \infty} (8 x - 5) x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2

Evaluate the limit of the numerator.

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Step 6.1.2.1

Apply the distributive property.

$\frac{lim_{x \to \infty} 8 x \cdot x - 5 x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2.2

Raise $x$ to the power of $1$ .

$\frac{lim_{x \to \infty} 8 (x^{1} x) - 5 x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2.3

Raise $x$ to the power of $1$ .

$\frac{lim_{x \to \infty} 8 (x^{1} x^{1}) - 5 x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{lim_{x \to \infty} 8 x^{1 + 1} - 5 x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2.5

Add $1$ and $1$ .

$\frac{lim_{x \to \infty} 8 x^{2} - 5 x}{lim_{x \to \infty} 1 + 5 e^{5 x} x}$

Step 6.1.2.6

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

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

Step 6.1.3

Evaluate the limit of the denominator.

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Step 6.1.3.1

Evaluate the limit.

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Step 6.1.3.1.1

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

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

Step 6.1.3.1.2

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

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

Step 6.1.3.1.3

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

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

Step 6.1.3.1.4

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

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

Step 6.1.3.2

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

$\frac{\infty}{1 + 5 \cdot \infty \cdot lim_{x \to \infty} x}$

Step 6.1.3.3

Evaluate the limit.

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Step 6.1.3.3.1

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

$\frac{\infty}{1 + 5 \cdot \infty \cdot \infty}$

Step 6.1.3.3.2

Simplify the answer.

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Step 6.1.3.3.2.1

Simplify each term.

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Step 6.1.3.3.2.1.1

A non-zero constant times infinity is infinity.

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

Step 6.1.3.3.2.1.2

Infinity times infinity is infinity.

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

Step 6.1.3.3.2.2

Infinity plus or minus a number is infinity.

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

Step 6.1.3.3.2.3

Infinity divided by infinity is undefined.

Undefined

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

Step 6.1.3.3.3

Infinity divided by infinity is undefined.

Undefined

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

Step 6.1.3.4

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

$lim_{x \to \infty} \frac{(8 x - 5) x}{1 + 5 e^{5 x} x} = lim_{x \to \infty} \frac{\frac{d}{d x} [(8 x - 5) x]}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [(8 x - 5) x]}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = 8 x - 5$ and $g (x) = x$ .

$lim_{x \to \infty} \frac{(8 x - 5) \frac{d}{d x} [x] + x \frac{d}{d x} [8 x - 5]}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.3

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

$lim_{x \to \infty} \frac{(8 x - 5) \cdot 1 + x \frac{d}{d x} [8 x - 5]}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.4

Multiply $8 x - 5$ by $1$ .

$lim_{x \to \infty} \frac{8 x - 5 + x \frac{d}{d x} [8 x - 5]}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.5

By the Sum Rule, the derivative of $8 x - 5$ with respect to $x$ is $\frac{d}{d x} [8 x] + \frac{d}{d x} [- 5]$ .

$lim_{x \to \infty} \frac{8 x - 5 + x (\frac{d}{d x} [8 x] + \frac{d}{d x} [- 5])}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.6

Since $8$ is constant with respect to $x$ , the derivative of $8 x$ with respect to $x$ is $8 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{8 x - 5 + x (8 \frac{d}{d x} [x] + \frac{d}{d x} [- 5])}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.7

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

$lim_{x \to \infty} \frac{8 x - 5 + x (8 \cdot 1 + \frac{d}{d x} [- 5])}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.8

Multiply $8$ by $1$ .

$lim_{x \to \infty} \frac{8 x - 5 + x (8 + \frac{d}{d x} [- 5])}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.9

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

$lim_{x \to \infty} \frac{8 x - 5 + x (8 + 0)}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.10

Add $8$ and $0$ .

$lim_{x \to \infty} \frac{8 x - 5 + x \cdot 8}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.11

Move $8$ to the left of $x$ .

$lim_{x \to \infty} \frac{8 x - 5 + 8 \cdot x}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.12

Add $8 x$ and $8 x$ .

$lim_{x \to \infty} \frac{16 x - 5}{\frac{d}{d x} [1 + 5 e^{5 x} x]}$

Step 6.3.13

By the Sum Rule, the derivative of $1 + 5 e^{5 x} x$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [5 e^{5 x} x]$ .

$lim_{x \to \infty} \frac{16 x - 5}{\frac{d}{d x} [1] + \frac{d}{d x} [5 e^{5 x} x]}$

Step 6.3.14

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

$lim_{x \to \infty} \frac{16 x - 5}{0 + \frac{d}{d x} [5 e^{5 x} x]}$

Step 6.3.15

Evaluate $\frac{d}{d x} [5 e^{5 x} x]$ .

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Step 6.3.15.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{5 x} x$ with respect to $x$ is $5 \frac{d}{d x} [e^{5 x} x]$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 \frac{d}{d x} [e^{5 x} x]}$

Step 6.3.15.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{5 x}$ and $g (x) = x$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \frac{d}{d x} [x] + x \frac{d}{d x} [e^{5 x}])}$

Step 6.3.15.3

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

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x \frac{d}{d x} [e^{5 x}])}$

Step 6.3.15.4

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 $g (x) = 5 x$ .

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Step 6.3.15.4.1

To apply the Chain Rule, set $u$ as $5 x$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [5 x]))}$

Step 6.3.15.4.2

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

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x (e^{u} \frac{d}{d x} [5 x]))}$

Step 6.3.15.4.3

Replace all occurrences of $u$ with $5 x$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x (e^{5 x} \frac{d}{d x} [5 x]))}$

Step 6.3.15.5

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x (e^{5 x} (5 \frac{d}{d x} [x])))}$

Step 6.3.15.6

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

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} \cdot 1 + x (e^{5 x} (5 \cdot 1)))}$

Step 6.3.15.7

Multiply $e^{5 x}$ by $1$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} + x (e^{5 x} (5 \cdot 1)))}$

Step 6.3.15.8

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} + x (e^{5 x} \cdot 5))}$

Step 6.3.15.9

Move $5$ to the left of $e^{5 x}$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 (e^{5 x} + x (5 e^{5 x}))}$

Step 6.3.16

Simplify.

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Step 6.3.16.1

Apply the distributive property.

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 e^{5 x} + 5 (x (5 e^{5 x}))}$

Step 6.3.16.2

Combine terms.

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Step 6.3.16.2.1

Multiply $5$ by $5$ .

$lim_{x \to \infty} \frac{16 x - 5}{0 + 5 e^{5 x} + 25 (x (e^{5 x}))}$

Step 6.3.16.2.2

Add $0$ and $5 e^{5 x} + 25 x e^{5 x}$ .

$lim_{x \to \infty} \frac{16 x - 5}{5 e^{5 x} + 25 x e^{5 x}}$

Step 6.3.16.3

Reorder terms.

$lim_{x \to \infty} \frac{16 x - 5}{25 e^{5 x} x + 5 e^{5 x}}$

Step 6.3.16.4

Reorder factors in $25 e^{5 x} x + 5 e^{5 x}$ .

$lim_{x \to \infty} \frac{16 x - 5}{25 x e^{5 x} + 5 e^{5 x}}$

Step 7

Apply L'Hospital's rule.

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Step 7.1

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

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Step 7.1.1

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

$\frac{lim_{x \to \infty} 16 x - 5}{lim_{x \to \infty} 25 x e^{5 x} + 5 e^{5 x}}$

Step 7.1.2

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

$\frac{\infty}{lim_{x \to \infty} 25 x e^{5 x} + 5 e^{5 x}}$

Step 7.1.3

Evaluate the limit of the denominator.

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Step 7.1.3.1

Evaluate the limit.

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Step 7.1.3.1.1

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

$\frac{\infty}{lim_{x \to \infty} 25 x e^{5 x} + lim_{x \to \infty} 5 e^{5 x}}$

Step 7.1.3.1.2

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

$\frac{\infty}{lim_{x \to \infty} 25 \cdot lim_{x \to \infty} x \cdot lim_{x \to \infty} e^{5 x} + lim_{x \to \infty} 5 e^{5 x}}$

Step 7.1.3.1.3

Evaluate the limit of $25$ which is constant as $x$ approaches $\infty$ .

$\frac{\infty}{25 \cdot lim_{x \to \infty} x \cdot lim_{x \to \infty} e^{5 x} + lim_{x \to \infty} 5 e^{5 x}}$

Step 7.1.3.1.4

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

$\frac{\infty}{25 \cdot \infty \cdot lim_{x \to \infty} e^{5 x} + lim_{x \to \infty} 5 e^{5 x}}$

Step 7.1.3.2

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

$\frac{\infty}{25 \cdot \infty \cdot \infty + lim_{x \to \infty} 5 e^{5 x}}$

Step 7.1.3.3

Since the function $e^{5 x}$ approaches $\infty$ , the positive constant $5$ times the function also approaches $\infty$ .

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Step 7.1.3.3.1

Consider the limit with the constant multiple $5$ removed.

$\frac{\infty}{25 \cdot \infty \cdot \infty + lim_{x \to \infty} e^{5 x}}$

Step 7.1.3.3.2

Since the exponent $5 x$ approaches $\infty$ , the quantity $e^{5 x}$ approaches $\infty$ .

$\frac{\infty}{25 \cdot \infty \cdot \infty + \infty}$

Step 7.1.3.4

Simplify the answer.

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Step 7.1.3.4.1

Simplify each term.

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Step 7.1.3.4.1.1

A non-zero constant times infinity is infinity.

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

Step 7.1.3.4.1.2

Infinity times infinity is infinity.

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

Step 7.1.3.4.2

Infinity plus infinity is infinity.

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

Step 7.1.3.4.3

Infinity divided by infinity is undefined.

Undefined

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

Step 7.1.3.5

Infinity divided by infinity is undefined.

Undefined

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

Step 7.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 7.2

$lim_{x \to \infty} \frac{16 x - 5}{25 x e^{5 x} + 5 e^{5 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [16 x - 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3

Find the derivative of the numerator and denominator.

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Step 7.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [16 x - 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.2

By the Sum Rule, the derivative of $16 x - 5$ with respect to $x$ is $\frac{d}{d x} [16 x] + \frac{d}{d x} [- 5]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [16 x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.3

Evaluate $\frac{d}{d x} [16 x]$ .

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Step 7.3.3.1

Since $16$ is constant with respect to $x$ , the derivative of $16 x$ with respect to $x$ is $16 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{16 \frac{d}{d x} [x] + \frac{d}{d x} [- 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.3.2

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

$lim_{x \to \infty} \frac{16 \cdot 1 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.3.3

Multiply $16$ by $1$ .

$lim_{x \to \infty} \frac{16 + \frac{d}{d x} [- 5]}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.4

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

$lim_{x \to \infty} \frac{16 + 0}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.5

Add $16$ and $0$ .

$lim_{x \to \infty} \frac{16}{\frac{d}{d x} [25 x e^{5 x} + 5 e^{5 x}]}$

Step 7.3.6

By the Sum Rule, the derivative of $25 x e^{5 x} + 5 e^{5 x}$ with respect to $x$ is $\frac{d}{d x} [25 x e^{5 x}] + \frac{d}{d x} [5 e^{5 x}]$ .

$lim_{x \to \infty} \frac{16}{\frac{d}{d x} [25 x e^{5 x}] + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7

Evaluate $\frac{d}{d x} [25 x e^{5 x}]$ .

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Step 7.3.7.1

Since $25$ is constant with respect to $x$ , the derivative of $25 x e^{5 x}$ with respect to $x$ is $25 \frac{d}{d x} [x e^{5 x}]$ .

$lim_{x \to \infty} \frac{16}{25 \frac{d}{d x} [x e^{5 x}] + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{5 x}$ .

$lim_{x \to \infty} \frac{16}{25 (x \frac{d}{d x} [e^{5 x}] + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.3

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 $g (x) = 5 x$ .

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Step 7.3.7.3.1

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

$lim_{x \to \infty} \frac{16}{25 (x (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [5 x]) + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$lim_{x \to \infty} \frac{16}{25 (x (e^{u_{1}} \frac{d}{d x} [5 x]) + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.3.3

Replace all occurrences of $u_{1}$ with $5 x$ .

$lim_{x \to \infty} \frac{16}{25 (x (e^{5 x} \frac{d}{d x} [5 x]) + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.4

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{16}{25 (x (e^{5 x} (5 \frac{d}{d x} [x])) + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.5

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

$lim_{x \to \infty} \frac{16}{25 (x (e^{5 x} (5 \cdot 1)) + e^{5 x} \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.6

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

$lim_{x \to \infty} \frac{16}{25 (x (e^{5 x} (5 \cdot 1)) + e^{5 x} \cdot 1) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.7

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{16}{25 (x (e^{5 x} \cdot 5) + e^{5 x} \cdot 1) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.8

Move $5$ to the left of $e^{5 x}$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 \cdot e^{5 x}) + e^{5 x} \cdot 1) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.7.9

Multiply $e^{5 x}$ by $1$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + \frac{d}{d x} [5 e^{5 x}]}$

Step 7.3.8

Evaluate $\frac{d}{d x} [5 e^{5 x}]$ .

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Step 7.3.8.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{5 x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{5 x}]$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 \frac{d}{d x} [e^{5 x}]}$

Step 7.3.8.2

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 $g (x) = 5 x$ .

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Step 7.3.8.2.1

To apply the Chain Rule, set $u_{2}$ as $5 x$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [5 x])}$

Step 7.3.8.2.2

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

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (e^{u_{2}} \frac{d}{d x} [5 x])}$

Step 7.3.8.2.3

Replace all occurrences of $u_{2}$ with $5 x$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (e^{5 x} \frac{d}{d x} [5 x])}$

Step 7.3.8.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (e^{5 x} (5 \frac{d}{d x} [x]))}$

Step 7.3.8.4

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

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (e^{5 x} (5 \cdot 1))}$

Step 7.3.8.5

Multiply $5$ by $1$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (e^{5 x} \cdot 5)}$

Step 7.3.8.6

Move $5$ to the left of $e^{5 x}$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 5 (5 \cdot e^{5 x})}$

Step 7.3.8.7

Multiply $5$ by $5$ .

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x}) + e^{5 x}) + 25 e^{5 x}}$

Step 7.3.9

Simplify.

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Step 7.3.9.1

Apply the distributive property.

$lim_{x \to \infty} \frac{16}{25 (x (5 e^{5 x})) + 25 e^{5 x} + 25 e^{5 x}}$

Step 7.3.9.2

Combine terms.

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Step 7.3.9.2.1

Multiply $5$ by $25$ .

$lim_{x \to \infty} \frac{16}{125 (x (e^{5 x})) + 25 e^{5 x} + 25 e^{5 x}}$

Step 7.3.9.2.2

Add $25 e^{5 x}$ and $25 e^{5 x}$ .

$lim_{x \to \infty} \frac{16}{125 x e^{5 x} + 50 e^{5 x}}$

Step 7.3.9.3

Reorder terms.

$lim_{x \to \infty} \frac{16}{125 e^{5 x} x + 50 e^{5 x}}$

Step 7.3.9.4

Reorder factors in $125 e^{5 x} x + 50 e^{5 x}$ .

$lim_{x \to \infty} \frac{16}{125 x e^{5 x} + 50 e^{5 x}}$

Step 8

Move the term $16$ outside of the limit because it is constant with respect to $x$ .

$16 lim_{x \to \infty} \frac{1}{125 x e^{5 x} + 50 e^{5 x}}$

Step 9

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{125 x e^{5 x} + 50 e^{5 x}}$ approaches $0$ .

$16 \cdot 0$

Step 10

Multiply $16$ by $0$ .

$0$