Let f(x)=sin x, g(x)=cos x, h(x)=x^2,then lim | vedclass

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(B) Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$. We need to evaluate the limit $L = \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$. This ex

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

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(B) Given $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$. We need to evaluate the limit $L = \lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}$. This expression is the definition of the derivative of the composite function $F(x) = f(g(h(x)))$ at $x=1$,i.e.,$F'(1)$. First,find $F(x) = f(g(h(x))) = \sin(\cos(x^2))$. Now,differentiate $F(x)$ with respect to $x$ using the chain rule: $F'(x) = \cos(\cos(x^2)) \cdot \frac{d}{dx}(\cos(x^2)) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot \frac{d}{dx}(x^2)$. $F'(x) = \cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot (2x) = -2x \sin(x^2) \cos(\cos(x^2))$. Now,evaluate at $x=1$: $F'(1) = -2(1) \sin(1^2) \cos(\cos(1^2)) = -2 \sin 1 \cos(\cos 1)$. Thus,the limit is $-2 \sin 1 \cos(\cos 1)$. Therefore,option $B$ is correct.

Let $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$,then $\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=$

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