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

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Question

(A) We need to evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}$.Using the Taylor series expansions for ${e^{{x^2}}}$

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(A) We need to evaluate $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}$.Using the Taylor series expansions for ${e^{{x^2}}}$ and $\cos x$ around $x = 0$:${e^{{x^2}}} = 1 + {x^2} + \frac{{{x^4}}}{{2!}} + \dots$$\cos x = 1 - \frac{{{x^2}}}{{2!}} + \frac{{{x^4}}}{{4!}} - \dots$Substituting these into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{{(1 + {x^2} + \frac{{{x^4}}}{2} + \dots) - (1 - \frac{{{x^2}}}{2} + \frac{{{x^4}}}{{24}} - \dots)}}{{{x^2}}}$$= \mathop {\lim }\limits_{x 	o 0} \frac{{{x^2} + \frac{{{x^2}}}{2} + O({x^4})}}{{{x^2}}}$$= \mathop {\lim }\limits_{x 	o 0} (1 + \frac{1}{2} + O({x^2})) = \frac{3}{2}$.Alternatively,using $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2x{e^{{x^2}}} + \sin x}}{{2x}}$$= \mathop {\lim }\limits_{x 	o 0} \left( {e^{{x^2}}} + \frac{{\sin x}}{{2x}} ight)$$= 1 + \frac{1}{2} = \frac{3}{2}$.

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

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