L'Hopital's Rule Math Example 2

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Example 2

hard
Find lim⁡x→∞xe−x\displaystyle\lim_{x \to \infty} x e^{-x}.

Solution

  1. 1
    As x→∞x \to \infty: x→∞x \to \infty and e−x→0e^{-x} \to 0, giving ∞⋅0\infty \cdot 0 form.
  2. 2
    Rewrite: xe−x=xexxe^{-x} = \frac{x}{e^x}, which is ∞∞\frac{\infty}{\infty}.
  3. 3
    Apply L'Hôpital: lim⁡x→∞xex=lim⁡x→∞1ex=0\lim_{x\to\infty}\frac{x}{e^x} = \lim_{x\to\infty}\frac{1}{e^x} = 0.

Answer

00
The form 0⋅∞0 \cdot \infty must be rewritten as 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} before applying L'Hôpital. Rewriting xe−x=x/exx e^{-x} = x/e^x gives ∞/∞\infty/\infty, then one application suffices.

About L'Hopital's Rule

If lim⁡x→af(x)g(x)\lim_{x \to a} \frac{f(x)}{g(x)} is an indeterminate form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, then lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} provided the right-hand limit exists (or is ±∞\pm\infty).

Learn more about L'Hopital's Rule →

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Example 4 medium

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