Calculus Examples

$\frac{\ln (7 x + 5)}{e^{7 x + 5}}$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{\ln (7 x + 5)}{e^{7 x + 5}}$ is undefined.

$x \leq - \frac{5}{7}$

Step 1.2

Since $\frac{\ln (7 x + 5)}{e^{7 x + 5}}$ $\to$ $\infty$ as $x$ $\to$ $- \frac{5}{7}$ from the left and $\frac{\ln (7 x + 5)}{e^{7 x + 5}}$ $\to$ $- \infty$ as $x$ $\to$ $- \frac{5}{7}$ from the right, then $x = - \frac{5}{7}$ is a vertical asymptote.

$x = - \frac{5}{7}$

Step 1.3

Evaluate $lim_{x \to \infty} \frac{\ln (7 x + 5)}{e^{7 x + 5}}$ to find the horizontal asymptote.

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Step 1.3.1

Apply L'Hospital's rule.

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Step 1.3.1.1

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

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Step 1.3.1.1.1

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

$\frac{lim_{x \to \infty} \ln (7 x + 5)}{lim_{x \to \infty} e^{7 x + 5}}$

Step 1.3.1.1.2

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

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

Step 1.3.1.1.3

Since the exponent $7 x + 5$ approaches $\infty$ , the quantity $e^{7 x + 5}$ approaches $\infty$ .

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

Step 1.3.1.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 1.3.1.2

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.

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

Step 1.3.1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.1.3.2

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 $g (x) = 7 x + 5$ .

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Step 1.3.1.3.2.1

To apply the Chain Rule, set $u$ as $7 x + 5$ .

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

Step 1.3.1.3.2.2

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

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

Step 1.3.1.3.2.3

Replace all occurrences of $u$ with $7 x + 5$ .

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

Step 1.3.1.3.3

By the Sum Rule, the derivative of $7 x + 5$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [5]$ .

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

Step 1.3.1.3.4

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

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

Step 1.3.1.3.5

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

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

Step 1.3.1.3.6

Multiply $7$ by $1$ .

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

Step 1.3.1.3.7

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

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

Step 1.3.1.3.8

Add $7$ and $0$ .

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

Step 1.3.1.3.9

Combine $\frac{1}{7 x + 5}$ and $7$ .

$lim_{x \to \infty} \frac{\frac{7}{7 x + 5}}{\frac{d}{d x} [e^{7 x + 5}]}$

Step 1.3.1.3.10

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 $g (x) = 7 x + 5$ .

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Step 1.3.1.3.10.1

To apply the Chain Rule, set $u$ as $7 x + 5$ .

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

Step 1.3.1.3.10.2

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

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

Step 1.3.1.3.10.3

Replace all occurrences of $u$ with $7 x + 5$ .

$lim_{x \to \infty} \frac{\frac{7}{7 x + 5}}{e^{7 x + 5} \frac{d}{d x} [7 x + 5]}$

Step 1.3.1.3.11

By the Sum Rule, the derivative of $7 x + 5$ with respect to $x$ is $\frac{d}{d x} [7 x] + \frac{d}{d x} [5]$ .

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

Step 1.3.1.3.12

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

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

Step 1.3.1.3.13

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

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

Step 1.3.1.3.14

Multiply $7$ by $1$ .

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

Step 1.3.1.3.15

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

$lim_{x \to \infty} \frac{\frac{7}{7 x + 5}}{e^{7 x + 5} (7 + 0)}$

Step 1.3.1.3.16

Add $7$ and $0$ .

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

Step 1.3.1.3.17

Move $7$ to the left of $e^{7 x + 5}$ .

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

Step 1.3.1.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.3.1.5

Multiply $\frac{7}{7 x + 5}$ by $\frac{1}{7 e^{7 x + 5}}$ .

$lim_{x \to \infty} \frac{7}{(7 x + 5) (7 e^{7 x + 5})}$

Step 1.3.1.6

Cancel the common factor of $7$ .

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Step 1.3.1.6.1

Cancel the common factor.

$lim_{x \to \infty} \frac{7}{(7 x + 5) (7 e^{7 x + 5})}$

Step 1.3.1.6.2

Rewrite the expression.

$lim_{x \to \infty} \frac{1}{(7 x + 5) (e^{7 x + 5})}$

Step 1.3.2

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{(7 x + 5) (e^{7 x + 5})}$ approaches $0$ .

$0$

Step 1.4

List the horizontal asymptotes:

$y = 0$

Step 1.5

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.6

This is the set of all asymptotes.

Vertical Asymptotes: $x = - \frac{5}{7}$

Horizontal Asymptotes: $y = 0$

Vertical Asymptotes: $x = - \frac{5}{7}$

Horizontal Asymptotes: $y = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \frac{\ln (7 (1) + 5)}{e^{7 (1) + 5}}$

Step 2.2

Simplify the result.

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Step 2.2.1

Simplify the numerator.

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Step 2.2.1.1

Multiply $7$ by $1$ .

$f (1) = \frac{\ln (7 + 5)}{e^{7 (1) + 5}}$

Step 2.2.1.2

Add $7$ and $5$ .

$f (1) = \frac{\ln (12)}{e^{7 (1) + 5}}$

Step 2.2.2

Simplify the denominator.

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Step 2.2.2.1

Multiply $7$ by $1$ .

$f (1) = \frac{\ln (12)}{e^{7 + 5}}$

Step 2.2.2.2

Add $7$ and $5$ .

$f (1) = \frac{\ln (12)}{e^{12}}$

Step 2.2.3

The final answer is $\frac{\ln (12)}{e^{12}}$ .

$\frac{\ln (12)}{e^{12}}$

Step 2.3

Convert $\frac{\ln (12)}{e^{12}}$ to decimal.

$y = 1.52677941 \times 10^{- 5}$

Step 3

Find the point at $x = 2$ .

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Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \frac{\ln (7 (2) + 5)}{e^{7 (2) + 5}}$

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify the numerator.

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Step 3.2.1.1

Multiply $7$ by $2$ .

$f (2) = \frac{\ln (14 + 5)}{e^{7 (2) + 5}}$

Step 3.2.1.2

Add $14$ and $5$ .

$f (2) = \frac{\ln (19)}{e^{7 (2) + 5}}$

Step 3.2.2

Simplify the denominator.

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Step 3.2.2.1

Multiply $7$ by $2$ .

$f (2) = \frac{\ln (19)}{e^{14 + 5}}$

Step 3.2.2.2

Add $14$ and $5$ .

$f (2) = \frac{\ln (19)}{e^{19}}$

Step 3.2.3

The final answer is $\frac{\ln (19)}{e^{19}}$ .

$\frac{\ln (19)}{e^{19}}$

Step 3.3

Convert $\frac{\ln (19)}{e^{19}}$ to decimal.

$y = 1.64970922 \times 10^{- 8}$

Step 4

Find the point at $x = 3$ .

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Step 4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \frac{\ln (7 (3) + 5)}{e^{7 (3) + 5}}$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify the numerator.

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Step 4.2.1.1

Multiply $7$ by $3$ .

$f (3) = \frac{\ln (21 + 5)}{e^{7 (3) + 5}}$

Step 4.2.1.2

Add $21$ and $5$ .

$f (3) = \frac{\ln (26)}{e^{7 (3) + 5}}$

Step 4.2.2

Simplify the denominator.

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Step 4.2.2.1

Multiply $7$ by $3$ .

$f (3) = \frac{\ln (26)}{e^{21 + 5}}$

Step 4.2.2.2

Add $21$ and $5$ .

$f (3) = \frac{\ln (26)}{e^{26}}$

Step 4.2.3

The final answer is $\frac{\ln (26)}{e^{26}}$ .

$\frac{\ln (26)}{e^{26}}$

Step 4.3

Convert $\frac{\ln (26)}{e^{26}}$ to decimal.

$y = 1.66459052 \times 10^{- 11}$

Step 5

The log function can be graphed using the vertical asymptote at $x = - \frac{5}{7}$ and the points $(1, 1.52677941), (2, 1.64970922), (3, 1.66459052)$ .

Vertical Asymptote: $x = - \frac{5}{7}$

$\begin{array}{c} x & y \\ 1 & 1.527 \\ 2 & 1.65 \\ 3 & 1.665 \end{array}$

Step 6