A) \[\frac{2}{3}\]
B) \[\frac{1}{2}\]
C) \[\frac{3}{2}\]
D) 2
Correct Answer: C
Solution :
[c] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-\cos x}{{{x}^{2}}}\] which is \[\frac{0}{0}\]form \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{{{x}^{2}}}}.2x+\sin x}{2x}\] [By L' Hospital Rule] Which is \[\frac{0}{0}\]form \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{2.{{e}^{{{x}^{2}}}}+4{{x}^{2}}.{{e}^{{{x}^{2}}}}\cos x}{2}\] [By L' Hospital Rule] Applying limit\[x\to 0\], we have \[=\frac{2\times 1+0+1}{2}=\frac{3}{2}\]
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