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

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(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{e^{x^2} - \cos x}{x^2}$To evaluate this li

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$\frac{3}{2}$

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(D) Since $f(x)$ is continuous at $x = 0$,we have $f(0) = \lim_{x 	o 0} f(x)$.$f(0) = \lim_{x 	o 0} \frac{e^{x^2} - \cos x}{x^2}$To evaluate this limit,we add and subtract $1$ in the numerator:$f(0) = \lim_{x 	o 0} \frac{(e^{x^2} - 1) + (1 - \cos x)}{x^2}$$f(0) = \lim_{x 	o 0} \frac{e^{x^2} - 1}{x^2} + \lim_{x 	o 0} \frac{1 - \cos x}{x^2}$Using standard limits $\lim_{u 	o 0} \frac{e^u - 1}{u} = 1$ and $\lim_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$:$f(0) = 1 + \frac{1}{2} = \frac{3}{2}$.

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

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