Exercise · ZW-LHOPITALS-RULE-011 · ★☆☆ Basic
Compute \(\displaystyle\lim_{x\to0}\frac{e^{2x}-1}{x}\).
Choose one option.
Show answer
- C. \(2\)
Explanation
It has the form \(0/0\). Differentiating: \(\frac{2e^{2x}}{1}\to2\).
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-005 Compute lim _x→ ∞ x^2 e^- 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).