Question

The value of $\underset{h\to 0}{lim}\frac{(2+h)\cos (2+h)-2\cos 2}{h}$ is

• A. $\cos 2-2\sin 2$
• B. $\cos 2+2\sin 2$
• C. $\sin 2-2\cos 2$
• D. $\sin 2+2\cos 2$

Answer 1

The correct option is A $\cos 2-2\sin 2$

The given expression is equivalent to,

$\underset{x\to 2}{lim}\frac{f\left(x\right)-f\left(2\right)}{x-2},$ where $f\left(x\right)=x\cos x$

This is equal to $f^{\prime }\left(x\right)$ at $x=2$

If $f\left(x\right)=x\cos x$

$f^{\prime }\left(x\right)=\cos x-x\sin x$

At $x=2$

$f^{\prime }\left(2\right)=\cos 2-2\sin 2$

$\therefore \underset{h\to 0}{lim}\frac{(2+h)\cos (2+h)-2\cos 2}{h}=\cos 2-2\sin 2$