Question

Let \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = x^2 \). Then, evaluate: \[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} \]

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

Hint:

Using chain rule and L'Hôpital's rule simplifies limit calculations for composite functions.

Answer 1

The Correct Option is B

Provided: \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = x^2 \). We need to evaluate the limit: \[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} \] Substitute \( h(x) = x^2 \): \[ \lim\limits_{x \to 1} \frac{\sin(\cos(x^2)) - \sin(\cos 1)}{x - 1} \] Apply L'Hôpital's rule: \[ \lim\limits_{x \to 1} \frac{\frac{d}{dx}(\sin(\cos(x^2))) - \frac{d}{dx}(\sin(\cos 1))}{\frac{d}{dx}(x - 1)} = \lim\limits_{x \to 1} \frac{\cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x}{1} \] Evaluate at \( x = 1 \): \[ = \cos(\cos(1^2)) \cdot (-\sin(1^2)) \cdot 2(1) = -2\sin 1 \cos(\cos 1) \]