A) \[\frac{3}{2}\]
B) \[-\frac{1}{2}\]
C) 1
D) None of these
Correct Answer: A
Solution :
\[x=\frac{1}{t},\] Now expanding \[{{e}^{{{x}^{2}}}}\] and \[\cos x,\] we get \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\frac{3{{x}^{2}}}{2\,!}+{{x}^{4}}\,\left( \frac{1}{2\,!}-\frac{1}{4\,!} \right)+.......}{{{x}^{2}}}=\frac{3}{2}\] Aliter : Apply L-Hospital?s rule, \[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{2x{{e}^{{{x}^{2}}}}+\sin x}{2x}=\underset{x\to 0}{\mathop{\lim }}\,\,{{e}^{{{x}^{2}}}}+\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x}{2x}=1+\frac{1}{2}=\frac{3}{2}.\]
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