Precalculus Examples

$\frac{7 + 3 e^{3 x}}{4 - 8 e^{3 x}}$

Step 1

Find where the expression $\frac{7 + 3 e^{3 x}}{4 - 8 e^{3 x}}$ is undefined.

$x = \frac{\ln (\frac{1}{2})}{3}$

Step 2

Evaluate $lim_{x \to \infty} \frac{7 + 3 e^{3 x}}{4 - 8 e^{3 x}}$ to find the horizontal asymptote.

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Step 2.1

Apply L'Hospital's rule.

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Step 2.1.1

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

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Step 2.1.1.1

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

$\frac{lim_{x \to \infty} 7 + 3 e^{3 x}}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.2

Evaluate the limit of the numerator.

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Step 2.1.1.2.1

Evaluate the limit.

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Step 2.1.1.2.1.1

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

$\frac{lim_{x \to \infty} 7 + lim_{x \to \infty} 3 e^{3 x}}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.2.1.2

Evaluate the limit of $7$ which is constant as $x$ approaches $\infty$ .

$\frac{7 + lim_{x \to \infty} 3 e^{3 x}}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.2.2

Since the function $e^{3 x}$ approaches $\infty$ , the positive constant $3$ times the function also approaches $\infty$ .

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Step 2.1.1.2.2.1

Consider the limit with the constant multiple $3$ removed.

$\frac{7 + lim_{x \to \infty} e^{3 x}}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.2.2.2

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

$\frac{7 + \infty}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.2.3

Infinity plus or minus a number is infinity.

$\frac{\infty}{lim_{x \to \infty} 4 - 8 e^{3 x}}$

Step 2.1.1.3

Evaluate the limit of the denominator.

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Step 2.1.1.3.1

Evaluate the limit.

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Step 2.1.1.3.1.1

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

$\frac{\infty}{lim_{x \to \infty} 4 - lim_{x \to \infty} 8 e^{3 x}}$

Step 2.1.1.3.1.2

Evaluate the limit of $4$ which is constant as $x$ approaches $\infty$ .

$\frac{\infty}{4 - lim_{x \to \infty} 8 e^{3 x}}$

Step 2.1.1.3.2

Since the function $e^{3 x}$ approaches $\infty$ , the positive constant $8$ times the function also approaches $\infty$ .

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Step 2.1.1.3.2.1

Consider the limit with the constant multiple $8$ removed.

$\frac{\infty}{4 - lim_{x \to \infty} e^{3 x}}$

Step 2.1.1.3.2.2

Since the exponent $3 x$ approaches $\infty$ , the quantity $e^{3 x}$ approaches $\infty$ .

$\frac{\infty}{4 - \infty}$

Step 2.1.1.3.3

Simplify the answer.

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Step 2.1.1.3.3.1

A non-zero constant times infinity is infinity.

$\frac{\infty}{4 + - \infty}$

Step 2.1.1.3.3.2

Infinity plus or minus a number is infinity.

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

Step 2.1.1.3.3.3

Infinity divided by infinity is undefined.

Undefined

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

Step 2.1.1.3.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.1.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 2.1.2

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.

$lim_{x \to \infty} \frac{7 + 3 e^{3 x}}{4 - 8 e^{3 x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [7 + 3 e^{3 x}]}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3

Find the derivative of the numerator and denominator.

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Step 2.1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [7 + 3 e^{3 x}]}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.2

By the Sum Rule, the derivative of $7 + 3 e^{3 x}$ with respect to $x$ is $\frac{d}{d x} [7] + \frac{d}{d x} [3 e^{3 x}]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [7] + \frac{d}{d x} [3 e^{3 x}]}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{0 + \frac{d}{d x} [3 e^{3 x}]}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4

Evaluate $\frac{d}{d x} [3 e^{3 x}]$ .

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Step 2.1.3.4.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{3 x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{3 x}]$ .

$lim_{x \to \infty} \frac{0 + 3 \frac{d}{d x} [e^{3 x}]}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.2

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 $g (x) = 3 x$ .

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Step 2.1.3.4.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

$lim_{x \to \infty} \frac{0 + 3 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x])}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.2.2

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

$lim_{x \to \infty} \frac{0 + 3 (e^{u} \frac{d}{d x} [3 x])}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.2.3

Replace all occurrences of $u$ with $3 x$ .

$lim_{x \to \infty} \frac{0 + 3 (e^{3 x} \frac{d}{d x} [3 x])}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{0 + 3 (e^{3 x} (3 \frac{d}{d x} [x]))}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.4

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

$lim_{x \to \infty} \frac{0 + 3 (e^{3 x} (3 \cdot 1))}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.5

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{0 + 3 (e^{3 x} \cdot 3)}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.6

Move $3$ to the left of $e^{3 x}$ .

$lim_{x \to \infty} \frac{0 + 3 (3 \cdot e^{3 x})}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.4.7

Multiply $3$ by $3$ .

$lim_{x \to \infty} \frac{0 + 9 e^{3 x}}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.5

Add $0$ and $9 e^{3 x}$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{\frac{d}{d x} [4 - 8 e^{3 x}]}$

Step 2.1.3.6

By the Sum Rule, the derivative of $4 - 8 e^{3 x}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- 8 e^{3 x}]$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{\frac{d}{d x} [4] + \frac{d}{d x} [- 8 e^{3 x}]}$

Step 2.1.3.7

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 + \frac{d}{d x} [- 8 e^{3 x}]}$

Step 2.1.3.8

Evaluate $\frac{d}{d x} [- 8 e^{3 x}]$ .

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Step 2.1.3.8.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 e^{3 x}$ with respect to $x$ is $- 8 \frac{d}{d x} [e^{3 x}]$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 \frac{d}{d x} [e^{3 x}]}$

Step 2.1.3.8.2

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 $g (x) = 3 x$ .

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Step 2.1.3.8.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [3 x])}$

Step 2.1.3.8.2.2

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (e^{u} \frac{d}{d x} [3 x])}$

Step 2.1.3.8.2.3

Replace all occurrences of $u$ with $3 x$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (e^{3 x} \frac{d}{d x} [3 x])}$

Step 2.1.3.8.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (e^{3 x} (3 \frac{d}{d x} [x]))}$

Step 2.1.3.8.4

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

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (e^{3 x} (3 \cdot 1))}$

Step 2.1.3.8.5

Multiply $3$ by $1$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (e^{3 x} \cdot 3)}$

Step 2.1.3.8.6

Move $3$ to the left of $e^{3 x}$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 8 (3 \cdot e^{3 x})}$

Step 2.1.3.8.7

Multiply $3$ by $- 8$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{0 - 24 e^{3 x}}$

Step 2.1.3.9

Subtract $24 e^{3 x}$ from $0$ .

$lim_{x \to \infty} \frac{9 e^{3 x}}{- 24 e^{3 x}}$

Step 2.1.4

Reduce.

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Step 2.1.4.1

Cancel the common factor of $9$ and $- 24$ .

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Step 2.1.4.1.1

Factor $3$ out of $9 e^{3 x}$ .

$lim_{x \to \infty} \frac{3 (3 e^{3 x})}{- 24 e^{3 x}}$

Step 2.1.4.1.2

Cancel the common factors.

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Step 2.1.4.1.2.1

Factor $3$ out of $- 24 e^{3 x}$ .

$lim_{x \to \infty} \frac{3 (3 e^{3 x})}{3 (- 8 e^{3 x})}$

Step 2.1.4.1.2.2

Cancel the common factor.

$lim_{x \to \infty} \frac{3 (3 e^{3 x})}{3 (- 8 e^{3 x})}$

Step 2.1.4.1.2.3

Rewrite the expression.

$lim_{x \to \infty} \frac{3 e^{3 x}}{- 8 e^{3 x}}$

Step 2.1.4.2

Cancel the common factor of $e^{3 x}$ .

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Step 2.1.4.2.1

Cancel the common factor.

$lim_{x \to \infty} \frac{3 e^{3 x}}{- 8 e^{3 x}}$

Step 2.1.4.2.2

Rewrite the expression.

$lim_{x \to \infty} \frac{3}{- 8}$

Step 2.2

Evaluate the limit.

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Step 2.2.1

Evaluate the limit of $\frac{3}{- 8}$ which is constant as $x$ approaches $\infty$ .

$\frac{3}{- 8}$

Step 2.2.2

Move the negative in front of the fraction.

$- \frac{3}{8}$

Step 3

Evaluate $lim_{x \to - \infty} \frac{7 + 3 e^{3 x}}{4 - 8 e^{3 x}}$ to find the horizontal asymptote.

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Step 3.1

Evaluate the limit.

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Step 3.1.1

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

$\frac{lim_{x \to - \infty} 7 + 3 e^{3 x}}{lim_{x \to - \infty} 4 - 8 e^{3 x}}$

Step 3.1.2

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

$\frac{lim_{x \to - \infty} 7 + lim_{x \to - \infty} 3 e^{3 x}}{lim_{x \to - \infty} 4 - 8 e^{3 x}}$

Step 3.1.3

Evaluate the limit of $7$ which is constant as $x$ approaches $- \infty$ .

$\frac{7 + lim_{x \to - \infty} 3 e^{3 x}}{lim_{x \to - \infty} 4 - 8 e^{3 x}}$

Step 3.1.4

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

$\frac{7 + 3 lim_{x \to - \infty} e^{3 x}}{lim_{x \to - \infty} 4 - 8 e^{3 x}}$

Step 3.2

Since the exponent $3 x$ approaches $- \infty$ , the quantity $e^{3 x}$ approaches $0$ .

$\frac{7 + 3 \cdot 0}{lim_{x \to - \infty} 4 - 8 e^{3 x}}$

Step 3.3

Evaluate the limit.

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Step 3.3.1

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

$\frac{7 + 3 \cdot 0}{lim_{x \to - \infty} 4 - lim_{x \to - \infty} 8 e^{3 x}}$

Step 3.3.2

Evaluate the limit of $4$ which is constant as $x$ approaches $- \infty$ .

$\frac{7 + 3 \cdot 0}{4 - lim_{x \to - \infty} 8 e^{3 x}}$

Step 3.3.3

Move the term $8$ outside of the limit because it is constant with respect to $x$ .

$\frac{7 + 3 \cdot 0}{4 - 8 lim_{x \to - \infty} e^{3 x}}$

Step 3.4

Since the exponent $3 x$ approaches $- \infty$ , the quantity $e^{3 x}$ approaches $0$ .

$\frac{7 + 3 \cdot 0}{4 - 8 \cdot 0}$

Step 3.5

Simplify the answer.

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Step 3.5.1

Simplify the numerator.

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Step 3.5.1.1

Multiply $3$ by $0$ .

$\frac{7 + 0}{4 - 8 \cdot 0}$

Step 3.5.1.2

Add $7$ and $0$ .

$\frac{7}{4 - 8 \cdot 0}$

Step 3.5.2

Simplify the denominator.

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Step 3.5.2.1

Multiply $- 8$ by $0$ .

$\frac{7}{4 + 0}$

Step 3.5.2.2

Add $4$ and $0$ .

$\frac{7}{4}$

Step 4

List the horizontal asymptotes:

$y = - \frac{3}{8}, \frac{7}{4}$

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: $x = \frac{\ln (\frac{1}{2})}{3}$

Horizontal Asymptotes: $y = - \frac{3}{8}, \frac{7}{4}$

No Oblique Asymptotes

Step 7