mathop {lim }limits_{x to 0} frac{{{e^{{x^2}} | vedclass

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(C) We evaluate the limit using the series expansion of $e^{x^2}$ and $\cos x$ near $x = 0$:$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \dots$$\cos x = 1 -

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$\frac{3}{2}$

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(C) We evaluate the limit using the series expansion of $e^{x^2}$ and $\cos x$ near $x = 0$:$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \dots$$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$$\sin x \approx x$ as $x 	o 0$,so $\sin^2 x \approx x^2$.Substituting these into the expression:$\mathop {\lim }\limits_{x 	o 0} \frac{(1 + x^2 + \dots) - (1 - \frac{x^2}{2} + \dots)}{x^2}$$= \mathop {\lim }\limits_{x 	o 0} \frac{x^2 + \frac{x^2}{2}}{x^2} = \mathop {\lim }\limits_{x 	o 0} \frac{\frac{3}{2}x^2}{x^2} = \frac{3}{2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{{\sin }^2}x}}$ is equal to

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