Calculus Examples

$lim_{x \to \infty} \frac{\ln (x)}{- 2 e^{x}}$

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.

$\frac{lim_{x \to \infty} \ln (x)}{lim_{x \to \infty} - 2 e^{x}}$

Step 1.2

As log approaches infinity, the value goes to $\infty$ .

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

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.

$lim_{x \to \infty} \frac{\ln (x)}{- 2 e^{x}} = lim_{x \to \infty} \frac{\frac{d}{d x} [\ln (x)]}{\frac{d}{d x} [- 2 e^{x}]}$

Step 3

Find the derivative of the numerator and denominator.

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Step 3.1

Differentiate the numerator and denominator.

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

Step 3.2

The derivative of $\ln (x)$ with respect to $x$ is $\frac{1}{x}$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 4

Multiply the numerator by the reciprocal of the denominator.

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

Step 5

Evaluate the limit.

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Step 5.1

Multiply $\frac{1}{x}$ by $\frac{1}{- 2 e^{x}}$ .

$lim_{x \to \infty} \frac{1}{x (- 2 e^{x})}$

Step 5.2

Move the term $\frac{1}{- 2}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{- 2} lim_{x \to \infty} \frac{1}{x (e^{x})}$

Step 6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{x (e^{x})}$ approaches $0$ .

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

Step 7

Simplify the answer.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

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

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Step 7.2.1

Multiply $0$ by $- 1$ .

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

Step 7.2.2

Multiply $0$ by $\frac{1}{2}$ .

$0$