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

Given: \( f(x) = \sin x \), \( g(x) = \cos x \), and \( h(x) = x^2 \) \[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} \] Substituting \( h(x) = x^2 \) and evaluating at \( x = 1 \): \[ \lim\limits_{x \to 1} \frac{\sin(\cos(x^2)) - \sin(\cos 1)}{x - 1} \] Using L'Hôpital's rule: \[ \lim\limits_{x \to 1} \frac{\cos(\cos(x^2)) \cdot (-\sin(x^2)) \cdot 2x}{1} \] Evaluating at \( x = 1 \): \[ = -2\sin 1 \cos(\cos 1) \]

Answer 2

Step 1: Understand the given functions and expressions
We are given the following functions:
- \( f(x) = \sin x \),
- \( g(x) = \cos x \),
- \( h(x) = x^2 \).
We need to evaluate the limit:
\[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(h(1)))}{x - 1} \]
Step 2: Substitute \( h(1) \) into the expression
First, calculate \( h(1) \):
\[ h(1) = 1^2 = 1. \]
So, we now need to evaluate:
\[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - f(g(1))}{x - 1}. \] Since \( h(1) = 1 \), we can calculate \( g(1) \) and \( f(g(1)) \) next.

Step 3: Evaluate \( g(1) \) and \( f(g(1)) \)
We know that \( g(x) = \cos x \), so:
\[ g(1) = \cos(1). \]
Next, evaluate \( f(g(1)) \) using \( f(x) = \sin x \):
\[ f(g(1)) = \sin(\cos(1)). \]
Step 4: Approximate the limit using the chain rule
We now have:
\[ \lim\limits_{x \to 1} \frac{f(g(h(x))) - \sin(\cos(1))}{x - 1}. \] By applying the chain rule for derivatives, we need to find the derivative of the composite function \( f(g(h(x))) \) at \( x = 1 \). We start by differentiating each function in the composition:
- The derivative of \( f(x) = \sin x \) is \( f'(x) = \cos x \).
- The derivative of \( g(x) = \cos x \) is \( g'(x) = -\sin x \).
- The derivative of \( h(x) = x^2 \) is \( h'(x) = 2x \).
Using the chain rule, the derivative of \( f(g(h(x))) \) with respect to \( x \) is:
\[ \frac{d}{dx} f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) = \cos(g(h(x))) \cdot (-\sin(h(x))) \cdot 2x. \] Now, evaluate the derivative at \( x = 1 \):
- \( h(1) = 1 \),
- \( g(1) = \cos(1) \),
- \( f'(g(1)) = \cos(\cos(1)) \),
- \( h'(1) = 2 \).
Thus, at \( x = 1 \), the derivative is:
\[ \cos(\cos(1)) \cdot (-\sin(1)) \cdot 2 = -2 \sin(1) \cos(\cos(1)). \] Step 5: Final Answer
The value of the given limit is:

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