Doubt regarding a limit which is related to MVT

Let the function $f(x)$ be differentiable and $f'(x)$ be continuous in $\left(-\infty,\infty \right)$ with $f'(2)=14$ then evaluate the limit

$$\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}$$

My attempt:

$\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}$

$\lim_{x\to 0}\left(\frac{f(2+\sin x)-f(2+x\cos x)}{\sin x-x\cos x}\right)$ $\frac{\sin x-x\cos x}{x-\sin x}$$=\left(f'(2)\right)(2)=28$

Is the method used here correct

Solutions Collecting From Web of "Doubt regarding a limit which is related to MVT"

Your method is correct, but you need additional work to show that $$\lim_{x \to 0}\frac{\sin x – x\cos x}{x – \sin x} = 2$$ A possible approach is to use L’Hospital’s Rule or Taylor series. Also you should write that fraction $$\frac{f(2 + \sin x) – f(2 + x\cos x)}{\sin x – x\cos x}$$ is equal to $f'(c_{x})$ by mean value theorem where $c_{x}$ lies between $2 + \sin x$ and $2 + x\cos x$ and as $x \to 0$ the number $c_{x} \to 2$. By continuity of $f’$ the fraction tends to $f'(2)$.