Calculus Examples

lim x \to - \infty x e^{x}

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

Step 1

Rewrite

x e^{x}

$x e^{x}$ as

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

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

lim x \to - \infty \frac{x}{e^{- x}}

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

Step 2

Apply L'Hospital's rule.

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Step 2.1

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

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Step 2.1.1

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

\frac{lim x \to - \infty x}{lim x \to - \infty e^{- x}}

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

Step 2.1.2

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

\frac{- \infty}{lim x \to - \infty e^{- x}}

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

Step 2.1.3

Since the exponent

- x

$- x$ approaches

\infty

$\infty$ , the quantity

e^{- x}

$e^{- x}$ approaches

\infty

$\infty$ .

\frac{- \infty}{\infty}

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

Step 2.1.4

Infinity divided by infinity is undefined.

Undefined

\frac{- \infty}{\infty}

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

Step 2.2

Since

\frac{- \infty}{\infty}

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

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

Step 2.3

Find the derivative of the numerator and denominator.

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Step 2.3.1

Differentiate the numerator and denominator.

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

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

Step 2.3.2

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

lim x \to - \infty \frac{1}{\frac{d}{d x} [e^{- x}]}

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

Step 2.3.3

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

$f' (g (x)) g' (x)$ where

$f (x) = e^{x}$ and

$g (x) = - x$ .

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Step 2.3.3.1

To apply the Chain Rule, set

$u$ as

$- x$ .

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

Step 2.3.3.2

Differentiate using the Exponential Rule which states that

$\frac{d}{d u} [a^{u}]$ is

$a^{u} \ln (a)$ where

$a$ =

$e$ .

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

Step 2.3.3.3

Replace all occurrences of

$u$ with

$- x$ .

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

Step 2.3.4

Since

$- 1$ is constant with respect to

$x$ , the derivative of

$- x$ with respect to

$x$ is

$- \frac{d}{d x} [x]$ .

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

Step 2.3.5

Differentiate using the Power Rule which states that

$\frac{d}{d x} [x^{n}]$ is

$n x^{n - 1}$ where

$n = 1$ .

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

Step 2.3.6

Multiply

$- 1$ by

$1$ .

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

Step 2.3.7

Move

$- 1$ to the left of

$e^{- x}$ .

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

Step 2.3.8

Rewrite

$- 1 e^{- x}$ as

$- e^{- x}$ .

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

Step 2.4

Cancel the common factor of

$1$ and

$- 1$ .

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Step 2.4.1

Rewrite

$1$ as

$- 1 (- 1)$ .

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

Step 2.4.2

Move the negative in front of the fraction.

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

Step 3

Move the term

$- 1$ outside of the limit because it is constant with respect to

$x$ .

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

Step 4

Since its numerator approaches a real number while its denominator is unbounded, the fraction

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

$0$ .

$- 0$

Step 5

Multiply

$- 1$ by

$0$ .

$0$