Q107 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics
Let $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$ then $\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=$
- A0
- Correct answerB$-2 \sin 1 \cos (\cos 1)$
- C$\infty$
- D$-2 \sin 1 \cos 1$
Solution
Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$
$$
\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=\lim _{x \rightarrow 1} \frac{\sin \left(\cos x^2\right)-\sin (\cos 1)}{x-1}
$$
If Apply limit it gives $\frac{0}{0}$ form, then apply L'Hospital rule.
$$
\Rightarrow \lim _{x \rightarrow 1} \frac{\cos \left(\cos x^2\right)\left(-\sin x^2\right)(2 x)-0}{1-0}
$$
Apply the limit,
$$
\Rightarrow-2(1) \sin (1) \cos (\cos 1)=-2 \sin (1) \cos (\cos 1)
$$
So, option (b) is correct.
Hence, the correct answer is option: 'B'.