Question

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

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

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!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given limit is the definition of the derivative of the composite function \( F(x) = f(g(h(x))) \) at \( x = 1 \). According to the limit definition of a derivative: \[ F'(a) = \lim_{x \to a} \frac{F(x) - F(a)}{x - a} \]

Step 2: Key Formula or Approach:
We use the Chain Rule for differentiation: \[ \frac{d}{dx} [f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \]

Step 3: Detailed Explanation:
Given \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = x^2 \). Their derivatives are: \( f'(x) = \cos x \), \( g'(x) = -\sin x \), and \( h'(x) = 2x \). 1. Compute the general derivative \( F'(x) \): \[ F'(x) = \cos(g(h(x))) \cdot [-\sin(h(x))] \cdot [2x] \] 2. Substitute \( x = 1 \): \[ F'(1) = \cos(\cos(1^2)) \cdot [-\sin(1^2)] \cdot [2(1)] \] \[ F'(1) = \cos(\cos 1) \cdot (-\sin 1) \cdot 2 \] \[ F'(1) = -2 \sin 1 \cos(\cos 1) \]

Step 4: Final Answer
The value of the limit is \(-2 \sin 1 \cos(\cos 1)\).