Let f(x)= x, g(x)= x, h(x)=x^2 then _ x arrow 1 (f(g(h(x)))-f(g(h(1))))/(x-1)=

Correct answer: B. -2 1 ( 1) Given f(x)= x, g(x)= x, h(x)=x^2 _ x arrow 1 (f(g(h(x)))-f(g(h(1))))/(x-1) When limit it applied, it gives (0)/(0) form, so we apply L'Hospital rule. _ x arrow 1 ( ( x^2)(- x^2)(2 x)-0)/(1-0) Apply the limit, -2(1) (1) ( 1)=-2 (1) ( 1)

BITSAT 2024Paper 1

Exam: BITSAT
Year: 2024
Paper: Paper 1
Subject: Mathematics
Chapter: Limits

BITSAT 2024 · Paper 1 · Mathematics · Limits

Let f(x)= x, g(x)= x, h(x)=x^2 then _ x arrow 1...

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}=

Options

A $0$
B $-2 \sin 1 \cos (\cos 1)$
C $\infty$
D $-2 \sin 1 \cos 1$

Correct answer

B. $-2 \sin 1 \cos (\cos 1)$

Step-by-step 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}

When limit it applied, it gives

\frac{0}{0}

form, so we 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)

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