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

We are given the function: \[ f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} \]

and we are told that it is continuous at $x = 0$. For continuity, $f(0) = \lim_{x \to 0} f(x)$.

We use Taylor series expansions around $x=0$:

\[\begin{align} \tan x &= x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7) \\\sin x &= x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \end{align}\]

Step 1: Expansion of $\tan(\tan x)$

Let $u = \tan x = x + \frac{x^3}{3} + O(x^5)$.

\[\begin{align} \tan(\tan x) &= \tan(u) = u + \frac{u^3}{3} + O(u^5) \\&= \left(x + \frac{x^3}{3}\right) + \frac{1}{3}\left(x + \frac{x^3}{3}\right)^3 + O(x^5) \\&= \left(x + \frac{x^3}{3}\right) + \frac{1}{3}(x^3 + 3x^2(\frac{x^3}{3}) + ...) + O(x^5) \\&= \left(x + \frac{x^3}{3}\right) + \frac{1}{3}(x^3 + O(x^5)) + O(x^5) \\&= x + \frac{x^3}{3} + \frac{x^3}{3} + O(x^5) = x + \frac{2x^3}{3} + O(x^5) \end{align}\]

Step 2: Expansion of $\sin(\sin x)$

Let $v = \sin x = x - \frac{x^3}{6} + O(x^5)$.

\[\begin{align} \sin(\sin x) &= \sin(v) = v - \frac{v^3}{6} + O(v^5) \\&= \left(x - \frac{x^3}{6}\right) - \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3 + O(x^5) \\&= \left(x - \frac{x^3}{6}\right) - \frac{1}{6}(x^3 - 3x^2(\frac{x^3}{6}) + ...) + O(x^5) \\&= \left(x - \frac{x^3}{6}\right) - \frac{1}{6}(x^3 + O(x^5)) + O(x^5) \\&= x - \frac{x^3}{6} - \frac{x^3}{6} + O(x^5) = x - \frac{x^3}{3} + O(x^5) \end{align}\]

Step 3: Expansion of the denominator

\[\begin{align} \tan x - \sin x &= \left(x + \frac{x^3}{3} + O(x^5)\right) - \left(x - \frac{x^3}{6} + O(x^5)\right) \\&= x + \frac{x^3}{3} - x + \frac{x^3}{6} + O(x^5) \\&= \left(\frac{1}{3} + \frac{1}{6}\right)x^3 + O(x^5) = \frac{2+1}{6}x^3 + O(x^5) = \frac{x^3}{2} + O(x^5) \end{align}\]

Step 4: Finding the limit of $f(x)$ as $x \to 0$

\[\begin{align} \lim_{x \to 0} f(x) &= \lim_{x \to 0} \frac{(x + \frac{2x^3}{3} + O(x^5)) - (x - \frac{x^3}{3} + O(x^5))}{\frac{x^3}{2} + O(x^5)} \\&= \lim_{x \to 0} \frac{x + \frac{2x^3}{3} - x + \frac{x^3}{3} + O(x^5)}{\frac{x^3}{2} + O(x^5)} \\&= \lim_{x \to 0} \frac{x^3 + O(x^5)}{\frac{x^3}{2} + O(x^5)} \\&= \lim_{x \to 0} \frac{x^3(1 + O(x^2))}{x^3(\frac{1}{2} + O(x^2))} \\&= \frac{1}{\frac{1}{2}} = 2 \end{align}\]

Since $f(x)$ is continuous at $x=0$, $f(0) = \lim_{x \to 0} f(x) = 2$.

Final Answer:

The final answer is $2$