Calculus Examples

$\frac{x}{1 - \ln (x - 2)}$

Step 1

Find the asymptotes.

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Step 1.1

Find where the expression $\frac{x}{1 - \ln (x - 2)}$ is undefined.

$x \leq 2, x = e + 2$

Step 1.2

Since $\frac{x}{1 - \ln (x - 2)}$ $\to$ $\infty$ as $x$ $\to$ $e + 2$ from the left and $\frac{x}{1 - \ln (x - 2)}$ $\to$ $- \infty$ as $x$ $\to$ $e + 2$ from the right, then $x = e + 2$ is a vertical asymptote.

$x = e + 2$

Step 1.3

Since the limit does not exist, there are no horizontal asymptotes.

No Horizontal Asymptotes

Step 1.4

No oblique asymptotes are present for logarithmic and trigonometric functions.

No Oblique Asymptotes

Step 1.5

This is the set of all asymptotes.

Vertical Asymptotes: $x = e + 2$

No Horizontal Asymptotes

Vertical Asymptotes: $x = e + 2$

No Horizontal Asymptotes

Step 2

Find the point at $x = 6$ .

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Step 2.1

Replace the variable $x$ with $6$ in the expression.

$f (6) = \frac{6}{1 - \ln ((6) - 2)}$

Step 2.2

Simplify the result.

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Step 2.2.1

Subtract $2$ from $6$ .

$f (6) = \frac{6}{1 - \ln (4)}$

Step 2.2.2

The final answer is $\frac{6}{1 - \ln (4)}$ .

$\frac{6}{1 - \ln (4)}$

Step 2.3

Convert $\frac{6}{1 - \ln (4)}$ to decimal.

$y = - 15.53219669$

Step 3

Find the point at $x = 7$ .

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Step 3.1

Replace the variable $x$ with $7$ in the expression.

$f (7) = \frac{7}{1 - \ln ((7) - 2)}$

Step 3.2

Simplify the result.

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Step 3.2.1

Subtract $2$ from $7$ .

$f (7) = \frac{7}{1 - \ln (5)}$

Step 3.2.2

The final answer is $\frac{7}{1 - \ln (5)}$ .

$\frac{7}{1 - \ln (5)}$

Step 3.3

Convert $\frac{7}{1 - \ln (5)}$ to decimal.

$y = - 11.48599366$

Step 4

Find the point at $x = 8$ .

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Step 4.1

Replace the variable $x$ with $8$ in the expression.

$f (8) = \frac{8}{1 - \ln ((8) - 2)}$

Step 4.2

Simplify the result.

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Step 4.2.1

Subtract $2$ from $8$ .

$f (8) = \frac{8}{1 - \ln (6)}$

Step 4.2.2

The final answer is $\frac{8}{1 - \ln (6)}$ .

$\frac{8}{1 - \ln (6)}$

Step 4.3

Convert $\frac{8}{1 - \ln (6)}$ to decimal.

$y = - 10.10407871$

Step 5

The log function can be graphed using the vertical asymptote at $x = e + 2$ and the points $(6, - 15.53219669), (7, - 11.48599366), (8, - 10.10407871)$ .

Vertical Asymptote: $x = e + 2$

$\begin{array}{c} x & y \\ 6 & - 15.532 \\ 7 & - 11.486 \\ 8 & - 10.104 \end{array}$

Step 6