Calculus Examples

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

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.1.3

The limit at negative infinity of a polynomial of odd degree whose leading coefficient is positive is negative infinity.

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

Step 1.1.4

Infinity divided by infinity is undefined.

Undefined

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

Step 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.

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

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

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

Step 1.3.2

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 $g (x) = - x$ .

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Step 1.3.2.1

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

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

Step 1.3.2.2

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

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

Step 1.3.2.3

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

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

Step 1.3.3

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

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

Step 1.3.4

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

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

Step 1.3.5

Multiply $- 1$ by $1$ .

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

Step 1.3.6

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

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

Step 1.3.7

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

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

Step 1.3.8

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

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

Step 1.4

Divide $- e^{- x}$ by $1$ .

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

Step 2

Since the function $e^{- x}$ approaches $\infty$ , the negative constant $- 1$ times the function approaches $- \infty$ .

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Step 2.1

Consider the limit with the constant multiple $- 1$ removed.

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

Step 2.2

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

$\infty$

Step 2.3

Since the function $e^{- x}$ approaches $\infty$ , the negative constant $- 1$ times the function approaches $- \infty$ .

$- \infty$