Question:
\(\lim_{x\to 0}\frac{1}{x^{2}}\left(\mathrm{e}^{x^{2}}-\cos\frac{x}{2}\right)\) is equal to :
\(\lim_{x\to 0}\frac{1}{x^{2}}\left(\mathrm{e}^{x^{2}}-\cos\frac{x}{2}\right)\) is equal to :
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Split the limit and use \(\frac{e^{t}-1}{t}\to1\) and \(1-\cos u=2\sin^2\frac u2\).
Updated On: Nov 12, 2026
- \(\frac{3}{2}\)
- \(\frac{5}{8}\)
- \(\frac{9}{8}\)
- \(\frac{9}{4}\)
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The Correct Option is C
Solution and Explanation
Write \(\frac{e^{x^2}-\cos(x/2)}{x^2}=\frac{e^{x^2}-1}{x^2}+\frac{1-\cos(x/2)}{x^2}\).
First part: \(\frac{e^{x^2}-1}{x^2}\to1\).
Second part: \(\frac{1-\cos(x/2)}{x^2}=\frac{2\sin^2(x/4)}{x^2}=\frac{2}{16}\cdot\left(\frac{\sin(x/4)}{x/4}\right)^2\to\frac18\).
Limit \(=1+\frac18=\frac98\).
Answer: (3) \(\frac{9}{8}\)
First part: \(\frac{e^{x^2}-1}{x^2}\to1\).
Second part: \(\frac{1-\cos(x/2)}{x^2}=\frac{2\sin^2(x/4)}{x^2}=\frac{2}{16}\cdot\left(\frac{\sin(x/4)}{x/4}\right)^2\to\frac18\).
Limit \(=1+\frac18=\frac98\).
Answer: (3) \(\frac{9}{8}\)
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