Question

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

• A. $\frac{2}{3}$
• B. $\frac{5}{2}$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

Answer: (D)

$\frac{3}{2}$

f(x) is continous at $x=0$

So,$f\left(0\right)=\underset{x\to 0}{lim}\frac{e^{x^2}-\cos x}{x^2}$

$=\underset{x\to 0}{lim}\frac{e^{x^2}-1+1-\cos x}{x^2}$

$=\underset{x\to 0}{lim}\frac{(e^{x^2}-1)+(1-\cos x)}{x^2}$

$=\underset{x\to 0}{lim}\frac{(e^{x^2}-1)}{x^2}+\underset{x\to 0}{lim}\frac{1-\cos x}{x^2}$

$=\underset{x\to 0}{lim}\frac{(e^{x^2}-1)}{x^2}+\underset{x\to 0}{lim}\frac{2si{n}^2\left(\frac{x}{2}\right)}{x^2}$

$=1+\frac{1}{2}$

$=\frac{3}{2}$

So, answer is option (D)