Question

If the function $f\left(x\right)=\frac{e^{x^2}-\cos x}{x^2}$ for $x\ne 0$ continuous at $x=0$ then $f\left(0\right)=$

• A. $\frac{1}{2}$
• B. $\frac{3}{2}$
• C. $2$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{3}{2}$

Here , function is continuous ,
So, left limit and right limit boh are equal and equals to the value of $f\left(x\right)$ at $x=0$
Therefore
$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{e^{x^2}-cosx}{x^2}$
Solving limit by derivative approach,
=> $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{e^{x^2}.2x+sinx}{2x}$
=> $\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2e^{x^2}+e^{x^2}.x^2+cosx}{2x}$
Putting $x=0$, we get
=> $f\left(0\right)=\frac{2+0+1}{2}=\frac{3}{2}$