Calculus Examples

$lim_{x \to \infty} \frac{\ln (3 x + 5)}{\ln (7 x + 3) + 1}$

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.

$\frac{lim_{x \to \infty} \ln (3 x + 5)}{lim_{x \to \infty} \ln (7 x + 3) + 1}$

Step 1.2

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

$\frac{\infty}{lim_{x \to \infty} \ln (7 x + 3) + 1}$

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 $\infty$ .

$\frac{\infty}{lim_{x \to \infty} \ln (7 x + 3) + lim_{x \to \infty} 1}$

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Infinity plus or minus a number 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.

$lim_{x \to \infty} \frac{\ln (3 x + 5)}{\ln (7 x + 3) + 1} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (3 x + 5)]}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (3 x + 5)]}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

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 $g (x) = 3 x + 5$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $3 x + 5$ .

$lim_{x \to \infty} \frac{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [3 x + 5]}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.2.2

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

$lim_{x \to \infty} \frac{\frac{1}{u} \frac{d}{d x} [3 x + 5]}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.2.3

Replace all occurrences of $u$ with $3 x + 5$ .

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} \frac{d}{d x} [3 x + 5]}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.3

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

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} (\frac{d}{d x} [3 x] + \frac{d}{d x} [5])}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.4

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

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} (3 \frac{d}{d x} [x] + \frac{d}{d x} [5])}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.5

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

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} (3 \cdot 1 + \frac{d}{d x} [5])}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.6

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} (3 + \frac{d}{d x} [5])}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.7

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

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} (3 + 0)}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.8

Add $3$ and $0$ .

$lim_{x \to \infty} \frac{\frac{1}{3 x + 5} \cdot 3}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.9

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

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{d}{d x} [\ln (7 x + 3) + 1]}$

Step 3.10

By the Sum Rule, the derivative of $\ln (7 x + 3) + 1$ with respect to $x$ is $\frac{d}{d x} [\ln (7 x + 3)] + \frac{d}{d x} [1]$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{d}{d x} [\ln (7 x + 3)] + \frac{d}{d x} [1]}$

Step 3.11

Evaluate $\frac{d}{d x} [\ln (7 x + 3)]$ .

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Step 3.11.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 7 x + 3$ .

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Step 3.11.1.1

To apply the Chain Rule, set $u$ as $7 x + 3$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{d}{d u} [\ln (u)] \frac{d}{d x} [7 x + 3] + \frac{d}{d x} [1]}$

Step 3.11.1.2

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

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{u} \frac{d}{d x} [7 x + 3] + \frac{d}{d x} [1]}$

Step 3.11.1.3

Replace all occurrences of $u$ with $7 x + 3$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} \frac{d}{d x} [7 x + 3] + \frac{d}{d x} [1]}$

Step 3.11.2

By the Sum Rule, the derivative of $7 x + 3$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [3]$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} (\frac{d}{d x} [7 x] + \frac{d}{d x} [3]) + \frac{d}{d x} [1]}$

Step 3.11.3

Since $7$ is constant with respect to $x$ , the derivative of $7 x$ with respect to $x$ is $7 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} (7 \frac{d}{d x} [x] + \frac{d}{d x} [3]) + \frac{d}{d x} [1]}$

Step 3.11.4

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

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} (7 \cdot 1 + \frac{d}{d x} [3]) + \frac{d}{d x} [1]}$

Step 3.11.5

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} (7 \cdot 1 + 0) + \frac{d}{d x} [1]}$

Step 3.11.6

Multiply $7$ by $1$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} (7 + 0) + \frac{d}{d x} [1]}$

Step 3.11.7

Add $7$ and $0$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{1}{7 x + 3} \cdot 7 + \frac{d}{d x} [1]}$

Step 3.11.8

Combine $\frac{1}{7 x + 3}$ and $7$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{7}{7 x + 3} + \frac{d}{d x} [1]}$

Step 3.12

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

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{7}{7 x + 3} + 0}$

Step 3.13

Add $\frac{7}{7 x + 3}$ and $0$ .

$lim_{x \to \infty} \frac{\frac{3}{3 x + 5}}{\frac{7}{7 x + 3}}$

Step 4

Multiply the numerator by the reciprocal of the denominator.

$lim_{x \to \infty} \frac{3}{3 x + 5} \cdot \frac{7 x + 3}{7}$

Step 5

Evaluate the limit.

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Step 5.1

Multiply $\frac{3}{3 x + 5}$ by $\frac{7 x + 3}{7}$ .

$lim_{x \to \infty} \frac{3 (7 x + 3)}{(3 x + 5) \cdot 7}$

Step 5.2

Move the term $\frac{3}{7}$ outside of the limit because it is constant with respect to $x$ .

$\frac{3}{7} lim_{x \to \infty} \frac{7 x + 3}{3 x + 5}$

Step 6

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

$\frac{3}{7} lim_{x \to \infty} \frac{\frac{7 x}{x} + \frac{3}{x}}{\frac{3 x}{x} + \frac{5}{x}}$

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.

$\frac{3}{7} lim_{x \to \infty} \frac{\frac{7 x}{x} + \frac{3}{x}}{\frac{3 x}{x} + \frac{5}{x}}$

Step 7.1.2

Divide $7$ by $1$ .

$\frac{3}{7} lim_{x \to \infty} \frac{7 + \frac{3}{x}}{\frac{3 x}{x} + \frac{5}{x}}$

Step 7.2

Cancel the common factor of $x$ .

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Step 7.2.1

Cancel the common factor.

$\frac{3}{7} lim_{x \to \infty} \frac{7 + \frac{3}{x}}{\frac{3 x}{x} + \frac{5}{x}}$

Step 7.2.2

Divide $3$ by $1$ .

$\frac{3}{7} lim_{x \to \infty} \frac{7 + \frac{3}{x}}{3 + \frac{5}{x}}$

Step 7.3

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

$\frac{3}{7} \cdot \frac{lim_{x \to \infty} 7 + \frac{3}{x}}{lim_{x \to \infty} 3 + \frac{5}{x}}$

Step 7.4

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

$\frac{3}{7} \cdot \frac{lim_{x \to \infty} 7 + lim_{x \to \infty} \frac{3}{x}}{lim_{x \to \infty} 3 + \frac{5}{x}}$

Step 7.5

Evaluate the limit of $7$ which is constant as $x$ approaches $\infty$ .

$\frac{3}{7} \cdot \frac{7 + lim_{x \to \infty} \frac{3}{x}}{lim_{x \to \infty} 3 + \frac{5}{x}}$

Step 7.6

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

$\frac{3}{7} \cdot \frac{7 + 3 lim_{x \to \infty} \frac{1}{x}}{lim_{x \to \infty} 3 + \frac{5}{x}}$

Step 8

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

$\frac{3}{7} \cdot \frac{7 + 3 \cdot 0}{lim_{x \to \infty} 3 + \frac{5}{x}}$

Step 9

Evaluate the limit.

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Step 9.1

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

$\frac{3}{7} \cdot \frac{7 + 3 \cdot 0}{lim_{x \to \infty} 3 + lim_{x \to \infty} \frac{5}{x}}$

Step 9.2

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

$\frac{3}{7} \cdot \frac{7 + 3 \cdot 0}{3 + lim_{x \to \infty} \frac{5}{x}}$

Step 9.3

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{3}{7} \cdot \frac{7 + 3 \cdot 0}{3 + 5 lim_{x \to \infty} \frac{1}{x}}$

Step 10

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

$\frac{3}{7} \cdot \frac{7 + 3 \cdot 0}{3 + 5 \cdot 0}$

Step 11

Simplify the answer.

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Step 11.1

Simplify the numerator.

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Step 11.1.1

Multiply $3$ by $0$ .

$\frac{3}{7} \cdot \frac{7 + 0}{3 + 5 \cdot 0}$

Step 11.1.2

Add $7$ and $0$ .

$\frac{3}{7} \cdot \frac{7}{3 + 5 \cdot 0}$

Step 11.2

Simplify the denominator.

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Step 11.2.1

Multiply $5$ by $0$ .

$\frac{3}{7} \cdot \frac{7}{3 + 0}$

Step 11.2.2

Add $3$ and $0$ .

$\frac{3}{7} \cdot \frac{7}{3}$

Step 11.3

Cancel the common factor of $3$ .

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Step 11.3.1

Cancel the common factor.

$\frac{3}{7} \cdot \frac{7}{3}$

Step 11.3.2

Rewrite the expression.

$\frac{1}{7} \cdot 7$

Step 11.4

Cancel the common factor of $7$ .

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Step 11.4.1

Cancel the common factor.

$\frac{1}{7} \cdot 7$

Step 11.4.2

Rewrite the expression.

$1$