If f(x) = frac{e^{x^2} - cos x}{x^2} for x ne | vedclass

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(B) For a function to be continuous at $x = 0$,the limit of the function as $x 	o 0$ must equal $f(0)$.$f(0) = \lim_{x 	o 0} \frac{e^{x^2} - \cos x}

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$\frac{3}{2}$

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(B) For a function to be continuous at $x = 0$,the limit of the function as $x 	o 0$ must equal $f(0)$.$f(0) = \lim_{x 	o 0} \frac{e^{x^2} - \cos x}{x^2}$Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \dots$ and $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$:$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \dots$$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$Substituting these into the limit:$f(0) = \lim_{x 	o 0} \frac{(1 + x^2 + \frac{x^4}{2} + \dots) - (1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots)}{x^2}$$f(0) = \lim_{x 	o 0} \frac{x^2 + \frac{x^2}{2} + O(x^4)}{x^2}$$f(0) = \lim_{x 	o 0} \frac{\frac{3}{2}x^2 + O(x^4)}{x^2} = \frac{3}{2}$.

If $f(x) = \frac{e^{x^2} - \cos x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then $f(0) = $.

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