Question:medium

Let \( f(x) = \sin x \), \( g(x) = \cos x \), \( h(x) = x^2 \) then
\[ \lim_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} = \]

Show Hint

Whenever you see a limit in the form \(\frac{\phi(x) - \phi(a)}{x-a}\), don't try to solve the limit using L'Hôpital's rule immediately; just find the derivative \(\phi'(a)\). It is much faster!
Updated On: Apr 15, 2026
  • 0
  • \(-2 \sin 1 \cos(\cos 1)\)
  • \(\infty\)
  • \(-2 \sin 1 \cos 1\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept: 
The given limit expression matches the precise limit definition of a derivative at a specific point. 
\[ \lim_{x\rightarrow a} \frac{F(x) - F(a)}{x - a} = F'(a) \] Step 2: Key Formula or Approach: 
Let $F(x) = f(g(h(x)))$. We need to find the derivative $F'(x)$ using the Chain Rule and evaluate it at $x = 1$. 
Step 3: Detailed Explanation: 
First, construct the composite function $F(x)$: \[ F(x) = f(g(x^2)) = f(\cos(x^2)) = \sin(\cos(x^2)) \] Now, differentiate $F(x)$ with respect to $x$: \[ F'(x) = \frac{d}{dx} \left[ \sin(\cos(x^2)) \right] \] Apply the Chain Rule sequentially (outside to inside): \[ F'(x) = \cos(\cos(x^2)) \cdot \frac{d}{dx}[\cos(x^2)] \] \[ F'(x) = \cos(\cos(x^2)) \cdot [-\sin(x^2)] \cdot \frac{d}{dx}[x^2] \] \[ F'(x) = \cos(\cos(x^2)) \cdot [-\sin(x^2)] \cdot (2x) \] \[ F'(x) = -2x \sin(x^2) \cos(\cos(x^2)) \] Finally, evaluate the derivative at $x = 1$: \[ F'(1) = -2(1) \sin(1^2) \cos(\cos(1^2)) \] \[ F'(1) = -2 \sin(1) \cos(\cos 1) \] Step 4: Final Answer: 
The limit is $-2 \sin 1 \cos(\cos 1)$.

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