Calculus Examples

$f (x) = \frac{2 e^{x} - 6}{e^{x} + 1}$

Step 1

Find where the expression $\frac{2 e^{x} - 6}{e^{x} + 1}$ is undefined.

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 2

The vertical asymptotes occur at areas of infinite discontinuity.

No Vertical Asymptotes

Step 3

Evaluate $lim_{x \to \infty} \frac{2 e^{x} - 6}{e^{x} + 1}$ to find the horizontal asymptote.

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Step 3.1

Apply L'Hospital's rule.

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Step 3.1.1

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

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Step 3.1.1.1

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

$\frac{lim_{x \to \infty} 2 e^{x} - 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2

Evaluate the limit of the numerator.

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Step 3.1.1.2.1

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

$\frac{lim_{x \to \infty} 2 e^{x} - lim_{x \to \infty} 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2.2

Since the function $e^{x}$ approaches $\infty$ , the positive constant $2$ times the function also approaches $\infty$ .

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Step 3.1.1.2.2.1

Consider the limit with the constant multiple $2$ removed.

$\frac{lim_{x \to \infty} e^{x} - lim_{x \to \infty} 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2.2.2

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

$\frac{\infty - lim_{x \to \infty} 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2.3

Evaluate the limit.

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Step 3.1.1.2.3.1

Evaluate the limit of $6$ which is constant as $x$ approaches $\infty$ .

$\frac{\infty - 1 \cdot 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2.3.2

Simplify the answer.

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Step 3.1.1.2.3.2.1

Multiply $- 1$ by $6$ .

$\frac{\infty - 6}{lim_{x \to \infty} e^{x} + 1}$

Step 3.1.1.2.3.2.2

Infinity plus or minus a number is infinity.

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

Step 3.1.1.3

Evaluate the limit of the denominator.

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Step 3.1.1.3.1

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

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

Step 3.1.1.3.2

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

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

Step 3.1.1.3.3

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

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

Step 3.1.1.3.4

Infinity plus or minus a number is infinity.

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

Step 3.1.1.3.5

Infinity divided by infinity is undefined.

Undefined

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

Step 3.1.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 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.

$lim_{x \to \infty} \frac{2 e^{x} - 6}{e^{x} + 1} = lim_{x \to \infty} \frac{\frac{d}{d x} [2 e^{x} - 6]}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3

Find the derivative of the numerator and denominator.

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Step 3.1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [2 e^{x} - 6]}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.2

By the Sum Rule, the derivative of $2 e^{x} - 6$ with respect to $x$ is $\frac{d}{d x} [2 e^{x}] + \frac{d}{d x} [- 6]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [2 e^{x}] + \frac{d}{d x} [- 6]}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.3

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

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Step 3.1.3.3.1

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

$lim_{x \to \infty} \frac{2 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [- 6]}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.3.2

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

$lim_{x \to \infty} \frac{2 e^{x} + \frac{d}{d x} [- 6]}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.4

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{2 e^{x} + 0}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.5

Add $2 e^{x}$ and $0$ .

$lim_{x \to \infty} \frac{2 e^{x}}{\frac{d}{d x} [e^{x} + 1]}$

Step 3.1.3.6

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

$lim_{x \to \infty} \frac{2 e^{x}}{\frac{d}{d x} [e^{x}] + \frac{d}{d x} [1]}$

Step 3.1.3.7

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

$lim_{x \to \infty} \frac{2 e^{x}}{e^{x} + \frac{d}{d x} [1]}$

Step 3.1.3.8

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

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

Step 3.1.3.9

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

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

Step 3.1.4

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

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Step 3.1.4.1

Cancel the common factor.

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

Step 3.1.4.2

Divide $2$ by $1$ .

$lim_{x \to \infty} 2$

Step 3.2

Evaluate the limit of $2$ which is constant as $x$ approaches $\infty$ .

$2$

Step 4

Evaluate $lim_{x \to - \infty} \frac{2 e^{x} - 6}{e^{x} + 1}$ to find the horizontal asymptote.

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Step 4.1

Evaluate the limit.

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Step 4.1.1

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

$\frac{lim_{x \to - \infty} 2 e^{x} - 6}{lim_{x \to - \infty} e^{x} + 1}$

Step 4.1.2

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

$\frac{lim_{x \to - \infty} 2 e^{x} - lim_{x \to - \infty} 6}{lim_{x \to - \infty} e^{x} + 1}$

Step 4.1.3

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

$\frac{2 lim_{x \to - \infty} e^{x} - lim_{x \to - \infty} 6}{lim_{x \to - \infty} e^{x} + 1}$

Step 4.2

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

$\frac{2 \cdot 0 - lim_{x \to - \infty} 6}{lim_{x \to - \infty} e^{x} + 1}$

Step 4.3

Evaluate the limit.

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Step 4.3.1

Evaluate the limit of $6$ which is constant as $x$ approaches $- \infty$ .

$\frac{2 \cdot 0 - 1 \cdot 6}{lim_{x \to - \infty} e^{x} + 1}$

Step 4.3.2

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

$\frac{2 \cdot 0 - 1 \cdot 6}{lim_{x \to - \infty} e^{x} + lim_{x \to - \infty} 1}$

Step 4.4

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

$\frac{2 \cdot 0 - 1 \cdot 6}{0 + lim_{x \to - \infty} 1}$

Step 4.5

Evaluate the limit.

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Step 4.5.1

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

$\frac{2 \cdot 0 - 1 \cdot 6}{0 + 1}$

Step 4.5.2

Simplify the answer.

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Step 4.5.2.1

Simplify the numerator.

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Step 4.5.2.1.1

Multiply $2$ by $0$ .

$\frac{0 - 1 \cdot 6}{0 + 1}$

Step 4.5.2.1.2

Multiply $- 1$ by $6$ .

$\frac{0 - 6}{0 + 1}$

Step 4.5.2.1.3

Subtract $6$ from $0$ .

$\frac{- 6}{0 + 1}$

Step 4.5.2.2

Add $0$ and $1$ .

$\frac{- 6}{1}$

Step 4.5.2.3

Divide $- 6$ by $1$ .

$- 6$

Step 5

List the horizontal asymptotes:

$y = 2, - 6$

Step 6

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 7

This is the set of all asymptotes.

No Vertical Asymptotes

Horizontal Asymptotes: $y = 2, - 6$

No Oblique Asymptotes

Step 8