Calculus Examples

$lim_{x \to \infty} \frac{(6 e^{x} + 5 e^{- x})}{7 e^{x} + 4 e}$

Step 1

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

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Step 1.1

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

$\frac{lim_{x \to \infty} 6 e^{x} + 5 e^{- x}}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2

Evaluate the limit of the numerator.

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Step 1.2.1

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

$\frac{lim_{x \to \infty} 6 e^{x} + lim_{x \to \infty} 5 e^{- x}}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.2

Since the function $e^{x}$ approaches $\infty$ , the positive constant $6$ times the function also approaches $\infty$ .

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Step 1.2.2.1

Consider the limit with the constant multiple $6$ removed.

$\frac{lim_{x \to \infty} e^{x} + lim_{x \to \infty} 5 e^{- x}}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.2.2

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

$\frac{\infty + lim_{x \to \infty} 5 e^{- x}}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.3

Move the term $5$ outside of the limit because it is constant with respect to $x$ .

$\frac{\infty + 5 lim_{x \to \infty} e^{- x}}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.4

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

$\frac{\infty + 5 \cdot 0}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.5

Simplify the answer.

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Step 1.2.5.1

Multiply $5$ by $0$ .

$\frac{\infty + 0}{lim_{x \to \infty} 7 e^{x} + 4 e}$

Step 1.2.5.2

Add $\infty$ and $0$ .

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

Step 1.3

Evaluate the limit of the denominator.

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Step 1.3.1

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

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

Step 1.3.2

Since the function $e^{x}$ approaches $\infty$ , the positive constant $7$ times the function also approaches $\infty$ .

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Step 1.3.2.1

Consider the limit with the constant multiple $7$ removed.

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

Step 1.3.2.2

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

$\frac{\infty}{\infty + lim_{x \to \infty} 4 e}$

Step 1.3.3

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

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

Step 1.3.4

Infinity plus or minus a number is infinity.

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

Step 1.3.5

Infinity divided by infinity is undefined.

Undefined

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

Step 1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 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.

$lim_{x \to \infty} \frac{(6 e^{x} + 5 e^{- x})}{7 e^{x} + 4 e} = lim_{x \to \infty} \frac{\frac{d}{d x} [6 e^{x} + 5 e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

$lim_{x \to \infty} \frac{\frac{d}{d x} [6 e^{x} + 5 e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.2

By the Sum Rule, the derivative of $6 e^{x} + 5 e^{- x}$ with respect to $x$ is $\frac{d}{d x} [6 e^{x}] + \frac{d}{d x} [5 e^{- x}]$ .

$lim_{x \to \infty} \frac{\frac{d}{d x} [6 e^{x}] + \frac{d}{d x} [5 e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.3

Evaluate $\frac{d}{d x} [6 e^{x}]$ .

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Step 3.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 e^{x}$ with respect to $x$ is $6 \frac{d}{d x} [e^{x}]$ .

$lim_{x \to \infty} \frac{6 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [5 e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.3.2

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

$lim_{x \to \infty} \frac{6 e^{x} + \frac{d}{d x} [5 e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4

Evaluate $\frac{d}{d x} [5 e^{- x}]$ .

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Step 3.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{- x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{- x}]$ .

$lim_{x \to \infty} \frac{6 e^{x} + 5 \frac{d}{d x} [e^{- x}]}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 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 $g (x) = - x$ .

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Step 3.4.2.1

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

$lim_{x \to \infty} \frac{6 e^{x} + 5 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x])}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.2.2

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

$lim_{x \to \infty} \frac{6 e^{x} + 5 (e^{u} \frac{d}{d x} [- x])}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.2.3

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

$lim_{x \to \infty} \frac{6 e^{x} + 5 (e^{- x} \frac{d}{d x} [- x])}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$lim_{x \to \infty} \frac{6 e^{x} + 5 (e^{- x} (- \frac{d}{d x} [x]))}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.4

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

$lim_{x \to \infty} \frac{6 e^{x} + 5 (e^{- x} (- 1 \cdot 1))}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.5

Multiply $- 1$ by $1$ .

$lim_{x \to \infty} \frac{6 e^{x} + 5 (e^{- x} \cdot - 1)}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.6

Move $- 1$ to the left of $e^{- x}$ .

$lim_{x \to \infty} \frac{6 e^{x} + 5 (- 1 e^{- x})}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$lim_{x \to \infty} \frac{6 e^{x} + 5 (- e^{- x})}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.4.8

Multiply $- 1$ by $5$ .

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{\frac{d}{d x} [7 e^{x} + 4 e]}$

Step 3.5

By the Sum Rule, the derivative of $7 e^{x} + 4 e$ with respect to $x$ is $\frac{d}{d x} [7 e^{x}] + \frac{d}{d x} [4 e]$ .

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{\frac{d}{d x} [7 e^{x}] + \frac{d}{d x} [4 e]}$

Step 3.6

Evaluate $\frac{d}{d x} [7 e^{x}]$ .

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Step 3.6.1

Since $7$ is constant with respect to $x$ , the derivative of $7 e^{x}$ with respect to $x$ is $7 \frac{d}{d x} [e^{x}]$ .

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{7 \frac{d}{d x} [e^{x}] + \frac{d}{d x} [4 e]}$

Step 3.6.2

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

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{7 e^{x} + \frac{d}{d x} [4 e]}$

Step 3.7

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

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{7 e^{x} + 0}$

Step 3.8

Add $7 e^{x}$ and $0$ .

$lim_{x \to \infty} \frac{6 e^{x} - 5 e^{- x}}{7 e^{x}}$