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If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$ if $x \neq 0$ is continuous at $x=0$, then $f(0)=$.
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Verified Answer
The correct answer is:
$\frac{3}{2}$
For continuity at $x=0, \lim _{x \rightarrow 0} f(x)=f(0)$
$$
\begin{aligned}
& \Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}=\lim _{x \rightarrow 0} \\
& \frac{\left(1+x^2+\frac{x^4}{2 !}+\ldots \ldots\right)-\left(1-\frac{x^2}{2 !}+\ldots . . .\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\frac{3}{2} x^2+\ldots . .}{x^2}=\frac{3}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}=\lim _{x \rightarrow 0} \\
& \frac{\left(1+x^2+\frac{x^4}{2 !}+\ldots \ldots\right)-\left(1-\frac{x^2}{2 !}+\ldots . . .\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\frac{3}{2} x^2+\ldots . .}{x^2}=\frac{3}{2}
\end{aligned}
$$
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