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

If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$ if $x \neq 0$ is continuous at $x=0$, then $f(0)=$.
  1. $\frac{1}{2}$
  2. $\frac{3}{2}$
  3. $\frac{2}{3}$
  4. $\frac{-3}{2}$

Solution

For continuity at $x=0, \lim _{x \rightarrow 0} f(x)=f(0)$ $\begin{aligned} & \Rightarrow f(0)=\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos x}{x^2}=\lim _{x \rightarrow 0} \\ & \frac{\left(1+x^2+\frac{x^4}{2 !}+\ldots \ldots\right)-\left(1-\frac{x^2}{2 !}+\ldots . . .\right)}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\frac{3}{2} x^2+\ldots . .}{x^2}=\frac{3}{2} \end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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