Calculus Examples

$\frac{e^{x}}{e^{x} - e^{- 1}}$

Step 1

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

$x = - 1$

Step 2

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

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Step 2.1

Simplify terms.

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Step 2.1.1

Simplify the limit argument.

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Step 2.1.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 2.1.1.2

Combine terms.

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Step 2.1.1.2.1

To write $e^{x}$ as a fraction with a common denominator, multiply by $\frac{e}{e}$ .

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

Step 2.1.1.2.2

Combine the numerators over the common denominator.

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

Step 2.1.2

Simplify the limit argument.

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Step 2.1.2.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.1.2.2

Combine factors.

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Step 2.1.2.2.1

Multiply $e^{x}$ by $e$ .

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Step 2.1.2.2.1.1

Raise $e$ to the power of $1$ .

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

Step 2.1.2.2.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 2.1.2.2.2

Combine $e^{x}$ and $\frac{e}{e^{x + 1} - 1}$ .

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

Step 2.1.2.2.3

Multiply $e^{x}$ by $e$ .

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Step 2.1.2.2.3.1

Raise $e$ to the power of $1$ .

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

Step 2.1.2.2.3.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

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.

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

Step 2.2.1.2

Since the exponent $x + 1$ approaches $\infty$ , the quantity $e^{x + 1}$ approaches $\infty$ .

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

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 $\infty$ .

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

Step 2.2.1.3.2

Since the exponent $x + 1$ approaches $\infty$ , the quantity $e^{x + 1}$ approaches $\infty$ .

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

Step 2.2.1.3.3

Evaluate the limit.

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Step 2.2.1.3.3.1

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

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

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 $1$ .

$\frac{\infty}{\infty - 1}$

Step 2.2.1.3.3.2.2

Infinity plus or minus a number is infinity.

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

Step 2.2.1.3.3.2.3

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2.1.3.3.3

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2.1.3.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.2.1.4

Infinity divided by infinity is undefined.

Undefined

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

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.

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

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.

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

Step 2.2.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x + 1$ .

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Step 2.2.3.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.2.3.2.2

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

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

Step 2.2.3.2.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 2.2.3.3

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

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

Step 2.2.3.4

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

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

Step 2.2.3.5

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

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

Step 2.2.3.6

Add $1$ and $0$ .

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

Step 2.2.3.7

Multiply $e^{x + 1}$ by $1$ .

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

Step 2.2.3.8

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

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

Step 2.2.3.9

Evaluate $\frac{d}{d x} [e^{x + 1}]$ .

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Step 2.2.3.9.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = x + 1$ .

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Step 2.2.3.9.1.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.2.3.9.1.2

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

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

Step 2.2.3.9.1.3

Replace all occurrences of $u$ with $x + 1$ .

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

Step 2.2.3.9.2

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

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

Step 2.2.3.9.3

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

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

Step 2.2.3.9.4

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

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

Step 2.2.3.9.5

Add $1$ and $0$ .

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

Step 2.2.3.9.6

Multiply $e^{x + 1}$ by $1$ .

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

Step 2.2.3.10

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

$lim_{x \to \infty} \frac{e^{x + 1}}{e^{x + 1} + 0}$

Step 2.2.3.11

Add $e^{x + 1}$ and $0$ .

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

Step 2.2.4

Cancel the common factor of $e^{x + 1}$ .

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Step 2.2.4.1

Cancel the common factor.

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

Step 2.2.4.2

Rewrite the expression.

$lim_{x \to \infty} 1$

Step 2.3

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

$1$

Step 3

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

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Step 3.1

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

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

Step 3.2

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

$\frac{0}{lim_{x \to - \infty} e^{x} - e^{- 1}}$

Step 3.3

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

$\frac{0}{lim_{x \to - \infty} e^{x} - lim_{x \to - \infty} e^{- 1}}$

Step 3.4

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

$\frac{0}{0 - lim_{x \to - \infty} e^{- 1}}$

Step 3.5

Evaluate the limit.

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Step 3.5.1

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

$\frac{0}{0 - e^{- 1}}$

Step 3.5.2

Simplify the answer.

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Step 3.5.2.1

Simplify the denominator.

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Step 3.5.2.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{0}{0 - \frac{1}{e}}$

Step 3.5.2.1.2

Subtract $\frac{1}{e}$ from $0$ .

$\frac{0}{- \frac{1}{e}}$

Step 3.5.2.2

Multiply the numerator by the reciprocal of the denominator.

$0 (- e)$

Step 3.5.2.3

Multiply $0 (- e)$ .

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Step 3.5.2.3.1

Multiply $- 1$ by $0$ .

$0 e$

Step 3.5.2.3.2

Multiply $0$ by $e$ .

$0$

Step 4

List the horizontal asymptotes:

$y = 1, 0$

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: $x = - 1$

Horizontal Asymptotes: $y = 1, 0$

No Oblique Asymptotes

Step 7