Calculus Examples

$f (x) = \frac{2 e^{x}}{e^{x} - 9}$

Step 1

Find where the expression $\frac{2 e^{x}}{e^{x} - 9}$ is undefined.

$x = \ln (9)$

Step 2

Evaluate $lim_{x \to \infty} \frac{2 e^{x}}{e^{x} - 9}$ to find the horizontal asymptote.

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Step 2.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to \infty} \frac{e^{x}}{e^{x} - 9}$

Step 2.2

Apply L'Hospital's rule.

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Step 2.2.1

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

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Step 2.2.1.1

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

$2 \frac{lim_{x \to \infty} e^{x}}{lim_{x \to \infty} e^{x} - 9}$

Step 2.2.1.2

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

$2 \frac{\infty}{lim_{x \to \infty} e^{x} - 9}$

Step 2.2.1.3

Evaluate the limit of the denominator.

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Step 2.2.1.3.1

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

$2 \frac{\infty}{lim_{x \to \infty} e^{x} - lim_{x \to \infty} 9}$

Step 2.2.1.3.2

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

$2 \frac{\infty}{\infty - lim_{x \to \infty} 9}$

Step 2.2.1.3.3

Evaluate the limit.

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Step 2.2.1.3.3.1

Evaluate the limit of $9$ which is constant as $x$ approaches $\infty$ .

$2 \frac{\infty}{\infty - 1 \cdot 9}$

Step 2.2.1.3.3.2

Simplify the answer.

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Step 2.2.1.3.3.2.1

Multiply $- 1$ by $9$ .

$2 \frac{\infty}{\infty - 9}$

Step 2.2.1.3.3.2.2

Infinity plus or minus a number is infinity.

$2 \frac{\infty}{\infty}$

Step 2.2.1.3.3.2.3

Infinity divided by infinity is undefined.

Undefined

$2 \frac{\infty}{\infty}$

Step 2.2.1.3.3.3

Infinity divided by infinity is undefined.

Undefined

$2 \frac{\infty}{\infty}$

Step 2.2.1.3.4

Infinity divided by infinity is undefined.

Undefined

$2 \frac{\infty}{\infty}$

Step 2.2.1.4

Infinity divided by infinity is undefined.

Undefined

$2 \frac{\infty}{\infty}$

Step 2.2.2

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.

$lim_{x \to \infty} \frac{e^{x}}{e^{x} - 9} = lim_{x \to \infty} \frac{\frac{d}{d x} [e^{x}]}{\frac{d}{d x} [e^{x} - 9]}$

Step 2.2.3

Find the derivative of the numerator and denominator.

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Step 2.2.3.1

Differentiate the numerator and denominator.

$2 lim_{x \to \infty} \frac{\frac{d}{d x} [e^{x}]}{\frac{d}{d x} [e^{x} - 9]}$

Step 2.2.3.2

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

$2 lim_{x \to \infty} \frac{e^{x}}{\frac{d}{d x} [e^{x} - 9]}$

Step 2.2.3.3

By the Sum Rule, the derivative of $e^{x} - 9$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 9]$ .

$2 lim_{x \to \infty} \frac{e^{x}}{\frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 9]}$

Step 2.2.3.4

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

$2 lim_{x \to \infty} \frac{e^{x}}{e^{x} + \frac{d}{d x} [- 9]}$

Step 2.2.3.5

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9$ with respect to $x$ is $0$ .

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

Step 2.2.3.6

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

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

Step 2.2.4

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

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Step 2.2.4.1

Cancel the common factor.

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

Step 2.2.4.2

Rewrite the expression.

$2 lim_{x \to \infty} 1$

Step 2.3

Evaluate the limit.

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Step 2.3.1

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

$2 \cdot 1$

Step 2.3.2

Multiply $2$ by $1$ .

$2$

Step 3

Evaluate $lim_{x \to - \infty} \frac{2 e^{x}}{e^{x} - 9}$ to find the horizontal asymptote.

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Step 3.1

Evaluate the limit.

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Step 3.1.1

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$2 lim_{x \to - \infty} \frac{e^{x}}{e^{x} - 9}$

Step 3.1.2

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

$2 \frac{lim_{x \to - \infty} e^{x}}{lim_{x \to - \infty} e^{x} - 9}$

Step 3.2

Since the exponent $x$ approaches $- \infty$ , the quantity $e^{x}$ approaches $0$ .

$2 \frac{0}{lim_{x \to - \infty} e^{x} - 9}$

Step 3.3

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

$2 \frac{0}{lim_{x \to - \infty} e^{x} - lim_{x \to - \infty} 9}$

Step 3.4

Since the exponent $x$ approaches $- \infty$ , the quantity $e^{x}$ approaches $0$ .

$2 \frac{0}{0 - lim_{x \to - \infty} 9}$

Step 3.5

Evaluate the limit.

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Step 3.5.1

Evaluate the limit of $9$ which is constant as $x$ approaches $- \infty$ .

$2 \frac{0}{0 - 1 \cdot 9}$

Step 3.5.2

Simplify the answer.

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Step 3.5.2.1

Cancel the common factor of $0$ and $0 - 1 \cdot 9$ .

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Step 3.5.2.1.1

Rewrite $0$ as $- 1 (0)$ .

$2 \frac{0}{- 1 (0) - 1 \cdot 9}$

Step 3.5.2.1.2

Factor $- 1$ out of $- 1 \cdot 9$ .

$2 \frac{0}{- 1 (0) - 1 (9)}$

Step 3.5.2.1.3

Factor $- 1$ out of $- 1 (0) - 1 (9)$ .

$2 \frac{0}{- 1 (0 + 9)}$

Step 3.5.2.1.4

Factor $9$ out of $0$ .

$2 \frac{9 (0)}{- 1 (0 + 9)}$

Step 3.5.2.1.5

Cancel the common factors.

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Step 3.5.2.1.5.1

Factor $9$ out of $- 1 (0 + 9)$ .

$2 \frac{9 (0)}{9 (- 1 (0 + 1))}$

Step 3.5.2.1.5.2

Cancel the common factor.

$2 \frac{9 \cdot 0}{9 (- 1 (0 + 1))}$

Step 3.5.2.1.5.3

Rewrite the expression.

$2 \frac{0}{- 1 (0 + 1)}$

Step 3.5.2.2

Add $0$ and $1$ .

$2 \frac{0}{- 1 \cdot 1}$

Step 3.5.2.3

Multiply $- 1$ by $1$ .

$2 (\frac{0}{- 1})$

Step 3.5.2.4

Divide $0$ by $- 1$ .

$2 \cdot 0$

Step 3.5.2.5

Multiply $2$ by $0$ .

$0$

Step 4

List the horizontal asymptotes:

$y = 2, 0$

Step 5

There is no oblique asymptote because the degree of the numerator is less than or equal to the degree of the denominator.

No Oblique Asymptotes

Step 6

This is the set of all asymptotes.

Vertical Asymptotes: $x = \ln (9)$

Horizontal Asymptotes: $y = 2, 0$

No Oblique Asymptotes

Step 7