Question

If $f\left(x\right)=\frac{{(a+x)}^2\sin (a+x)-a^2\sin a}{x},x\ne 0$, the value of $f\left(0\right)$ so that f is continuous at $x=0$ is

• A. $a^2\cos a+a\sin a$
• B. $a^2\cos a+2a\sin a$
• C. $2a^2\cos a+a\sin a$
• D. none of these

Answer 1

The correct option is B $a^2\cos a+2a\sin a$

To be continuous at $x=0$, the value of the function at this point must be equal to the limit of the function here.
So, $f\left(0\right)={lim}_{x\to 0}f\left(x\right)={lim}_{x\to 0}\frac{{(a+x)}^2\sin (a+x)-a^2\sin a}{x}={lim}_{x\to 0}\frac{(a^2+x^2+2ax)(\sin a\cos x+\cos a\sin x)-a^2\sin a}{x}={lim}_{x\to 0}(\frac{{-a}^2\sin a(1-\cos x)}{x}+\frac{a^2\cos a\sin x}{x}+x(\sin a\cos x+\cos a\sin x)+2a(\sin a\cos x+\cos a\sin x))=a^2\cos a+2a\sin a$
(Using ${lim}_{x\to 0}\frac{(1-\cos x)}{x}=0,{lim}_{x\to 0}\frac{\sin x}{x}=1$)