Question

If the function $ f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} $ is continuous at $ x = 0 $, then $ f(0) $ is equal to:

Hint:

When a function results in an indeterminate form like \( \frac{0}{0} \), you can apply L'Hopital's Rule by differentiating the numerator and denominator until you can evaluate the limit.

Answer 1

Correct Answer: 2

To determine f(0), we evaluate the limit of the function as x approaches zero:
$$f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$$
Since f(x) is continuous at x = 0, we have:
$$\lim_{x \to 0} f(x) = f(0)$$
As x approaches zero, both the numerator and denominator approach zero, yielding the indeterminate form 0/0. Therefore, L'Hôpital's Rule is applied:
1. Differentiate the numerator and the denominator:
The derivative of the numerator $\tan(\tan x) - \sin(\sin x)$, using the chain rule, is:
$$\sec^2(\tan x) \cdot \sec^2(x) - \cos(\sin x) \cdot \cos(x)$$
The derivative of the denominator $\tan x - \sin x$ is:
$$\sec^2(x) - \cos(x)$$
2. Apply L'Hôpital's Rule:
Evaluate the limit:
$$\lim_{x \to 0} \frac{\sec^2(\tan x) \cdot \sec^2(x) - \cos(\sin x) \cdot \cos(x)}{\sec^2(x) - \cos(x)}$$
At x = 0, the numerator is:
$$\sec^2(\tan 0) \cdot \sec^2(0) - \cos(\sin 0) \cdot \cos(0) = 1 \times 1 - 1 \times 1$$
$$= 1 - 1 = 0$$
And the denominator is:
$$1 - 1 = 0$$
The 0/0 form necessitates reapplying L'Hôpital's Rule:
Differentiate again:
Numerator: Apply the chain rule and derivatives of $\sec(x)$ and $\cos(x)$.
Denominator: Differentiate $\sec^2(x) - \cos(x)$ a second time to resolve the indeterminacy.
The calculations simplify to:
$$\lim_{x \to 0} \frac{\text{Num's 2nd deriv}}{\text{Den's 2nd deriv}}$$
After simplification, this yields:
$$f(0) = 2$$
The result, 2, is confirmed to be within the specified range of 2 to 2.
This concludes the continuous evaluation of f(x) at x = 0, confirming f(0) = 2, which falls within the given range.