Exercise · ZW-LHOPITALS-RULE-005 · ★★☆ Intermediate
Compute \(\lim_{x\to \infty}\,x^{2} e^{- x}\).
Choose one option.
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- C. \(0\)
Explanation
Substituting gives the form 0·∞, so it is rewritten as \(\frac{x^{2}}{e^{x}}\), an ∞/∞ form. Differentiating twice: \(\frac{2x}{e^{x}}\) → \(\frac{2}{e^{x}}\). The resulting limit is \(0\).
Review the concept: L'Hôpital's rule
Rule that computes limits of quotients in indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) by differentiating numerator and denominator separately.
\[\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\]
More practice problems: l'hôpital's rule
- ZW-LHOPITALS-RULE-001 Compute lim _x→ 0 (- x + e^x - 1)/(x^2).
- ZW-LHOPITALS-RULE-002 Compute lim _x→ ∞ (ln (x ))/(x).
- ZW-LHOPITALS-RULE-003 Compute lim _x→ 0 (x - sin x)/(x^3).
- ZW-LHOPITALS-RULE-004 Compute lim _x→ 0 (e^2 x - 1)/(3 x).
- ZW-LHOPITALS-RULE-006 Compute lim _x→ 1 (ln (x ))/(x - 1).
- ZW-LHOPITALS-RULE-007 Compute lim _x→ 0 (sin (3 x ))/(x).
- ZW-LHOPITALS-RULE-008 Compute lim _x→ 2 (x^2 - 4)/(x - 2).
- ZW-LHOPITALS-RULE-009 Compute lim _x→ 0 (1 - cos x)/(x^2).