Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at $x=a$ if $\lim_{x \to a} f(x) = f(a)$.
• Standard Limits:
  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{\tan x}{x} = 1$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
• Taylor Series Expansions (around $x=0$):
  • $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
  • $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$

Step-by-Step Solution

The function is given by $f(x)=\frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x}$. For the function to be continuous at $x=0$, we must have $f(0) = \lim_{x \to 0} f(x)$. We need to evaluate this limit.

Step 1: Analyze the structure of the function and identify the need for limit evaluation. The function is undefined at $x=0$ in its current form because both the numerator and the denominator become zero (i.e., $\tan(\tan 0) - \sin(\sin 0) = \tan(0) - \sin(0) = 0 - 0 = 0$ and $\tan 0 - \sin 0 = 0 - 0 = 0$). This is an indeterminate form $\frac{0}{0}$, which requires limit evaluation.

Step 2: Simplify the denominator using standard limits or Taylor series. We know that $\tan x = x + \frac{x^3}{3} + O(x^5)$ and $\sin x = x - \frac{x^3}{6} + O(x^5)$. Therefore, $\tan x - \sin x = (x + \frac{x^3}{3}) - (x - \frac{x^3}{6}) + O(x^5) = \frac{x^3}{3} + \frac{x^3}{6} + O(x^5) = \frac{x^3}{2} + O(x^5)$. So, $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{2} + O(x^5)}{x^3} = \frac{1}{2}$.

Step 3: Simplify the numerator using Taylor series expansions. Let $y = \tan x$. As $x \to 0$, $y \to 0$. We use the Taylor expansion for $\tan y$ around $y=0$: $\tan y = y + \frac{y^3}{3} + O(y^5)$. Substituting $y = \tan x = x + \frac{x^3}{3} + O(x^5)$: $\tan(\tan x) = \tan(x + \frac{x^3}{3} + O(x^5))$ $= (x + \frac{x^3}{3} + O(x^5)) + \frac{1}{3}(x + \frac{x^3}{3} + O(x^5))^3 + O((x + \frac{x^3}{3} + O(x^5))^5)$ $= x + \frac{x^3}{3} + \frac{1}{3}(x^3 + O(x^5)) + O(x^5)$ $= x + \frac{x^3}{3} + \frac{x^3}{3} + O(x^5) = x + \frac{2x^3}{3} + O(x^5)$.

Now, consider the $\sin(\sin x)$ term. Let $z = \sin x$. As $x \to 0$, $z \to 0$. We use the Taylor expansion for $\sin z$ around $z=0$: $\sin z = z - \frac{z^3}{6} + O(z^5)$. Substituting $z = \sin x = x - \frac{x^3}{6} + O(x^5)$: $\sin(\sin x) = \sin(x - \frac{x^3}{6} + O(x^5))$ $= (x - \frac{x^3}{6} + O(x^5)) - \frac{1}{6}(x - \frac{x^3}{6} + O(x^5))^3 + O((x - \frac{x^3}{6} + O(x^5))^5)$ $= x - \frac{x^3}{6} - \frac{1}{6}(x^3 + O(x^5)) + O(x^5)$ $= x - \frac{x^3}{6} - \frac{x^3}{6} + O(x^5) = x - \frac{x^3}{3} + O(x^5)$.

Now, we can find the expansion of the numerator: $\tan(\tan x) - \sin(\sin x) = (x + \frac{2x^3}{3} + O(x^5)) - (x - \frac{x^3}{3} + O(x^5))$ $= x + \frac{2x^3}{3} - x + \frac{x^3}{3} + O(x^5) = x^3 + O(x^5)$.

Step 4: Evaluate the limit of the function. We want to find $\lim_{x \to 0} \frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x}$. Using the expansions from Step 2 and Step 3: $\lim_{x \to 0} \frac{x^3 + O(x^5)}{\frac{x^3}{2} + O(x^5)}$ Divide the numerator and denominator by $x^3$: $\lim_{x \to 0} \frac{1 + O(x^2)}{\frac{1}{2} + O(x^2)}$ As $x \to 0$, the terms $O(x^2)$ go to zero. So, the limit is $\frac{1}{\frac{1}{2}} = 2$.

Step 5: Relate the limit to the function's continuity. Since the function $f(x)$ is continuous at $x=0$, we have $f(0) = \lim_{x \to 0} f(x)$. Therefore, $f(0) = 2$.

Alternative Approach using Standard Limit Forms (Manipulating the expression):

Let's rewrite the expression by dividing the numerator and denominator by $x^3$: $f(x) = \frac{\frac{\tan (\tan x)-\sin (\sin x)}{x^3}}{\frac{\tan x-\sin x}{x^3}}$

We already found $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}$.

Now let's focus on the numerator $\lim_{x \to 0} \frac{\tan (\tan x)-\sin (\sin x)}{x^3}$. We can use the generalized limit $\lim_{u \to 0} \frac{\tan u - u}{u^3} = \frac{1}{3}$ and $\lim_{v \to 0} \frac{\sin v - v}{v^3} = -\frac{1}{6}$.

Let $u = \tan x$. Then $\frac{\tan(\tan x) - \tan x}{(\tan x)^3} \to \frac{1}{3}$ as $x \to 0$. And $\frac{\tan x - x}{x^3} \to \frac{1}{3}$ as $x \to 0$. So, $\tan(\tan x) - \tan x = \frac{1}{3}(\tan x)^3 + O(x^5) = \frac{1}{3}x^3 + O(x^5)$.

Let $v = \sin x$. Then $\frac{\sin(\sin x) - \sin x}{(\sin x)^3} \to -\frac{1}{6}$ as $x \to 0$. And $\frac{\sin x - x}{x^3} \to -\frac{1}{6}$ as $x \to 0$. So, $\sin(\sin x) - \sin x = -\frac{1}{6}(\sin x)^3 + O(x^5) = -\frac{1}{6}x^3 + O(x^5)$.

Therefore, $\tan(\tan x) - \sin(\sin x) = (\tan(\tan x) - \tan x) + (\tan x - \sin x) + (\sin x - \sin(\sin x))$ $= (\frac{1}{3}(\tan x)^3 + O(x^5)) + (\tan x - \sin x) + (-\frac{1}{6}(\sin x)^3 + O(x^5))$ $= \frac{1}{3}x^3 + O(x^5) + (\frac{x^3}{2} + O(x^5)) + (-\frac{1}{6}x^3 + O(x^5))$ $= x^3(\frac{1}{3} + \frac{1}{2} - \frac{1}{6}) + O(x^5)$ $= x^3(\frac{2+3-1}{6}) + O(x^5) = x^3(\frac{4}{6}) + O(x^5) = \frac{2}{3}x^3 + O(x^5)$.

Now, the limit of the numerator is $\lim_{x \to 0} \frac{\frac{2}{3}x^3 + O(x^5)}{x^3} = \frac{2}{3}$.

The limit of the function is $\frac{\lim_{x \to 0} \frac{\tan (\tan x)-\sin (\sin x)}{x^3}}{\lim_{x \to 0} \frac{\tan x-\sin x}{x^3}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{2}{3} \times 2 = \frac{4}{3}$.

Let's recheck the Taylor expansions. $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$ $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + \dots$

$\tan(\tan x) = \tan(x + \frac{x^3}{3} + \dots)$ $= (x + \frac{x^3}{3}) + \frac{1}{3}(x + \frac{x^3}{3})^3 + \dots$ $= x + \frac{x^3}{3} + \frac{1}{3}(x^3) + \dots = x + \frac{2x^3}{3} + \dots$

$\sin(\sin x) = \sin(x - \frac{x^3}{6} + \dots)$ $= (x - \frac{x^3}{6}) - \frac{1}{6}(x - \frac{x^3}{6})^3 + \dots$ $= x - \frac{x^3}{6} - \frac{1}{6}(x^3) + \dots = x - \frac{2x^3}{6} = x - \frac{x^3}{3} + \dots$

Numerator: $\tan(\tan x) - \sin(\sin x) = (x + \frac{2x^3}{3}) - (x - \frac{x^3}{3}) + \dots = \frac{3x^3}{3} + \dots = x^3 + \dots$

Denominator: $\tan x - \sin x = (x + \frac{x^3}{3}) - (x - \frac{x^3}{6}) + \dots = \frac{x^3}{3} + \frac{x^3}{6} + \dots = \frac{x^3}{2} + \dots$

So, $\lim_{x \to 0} \frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x} = \lim_{x \to 0} \frac{x^3 + \dots}{\frac{x^3}{2} + \dots} = \frac{1}{\frac{1}{2}} = 2$.

The provided solution's intermediate step seems to have a calculation error. Let's re-examine the standard limits for $\frac{\tan u - u}{u^3}$ and $\frac{\sin u - u}{u^3}$.

Using the generalized limit $\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}$. And $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}$.

Let's write the numerator as: $\tan(\tan x) - \sin(\sin x) = (\tan(\tan x) - \tan x) + (\tan x - \sin x) + (\sin x - \sin(\sin x))$

We have $\tan x - \sin x = x + \frac{x^3}{3} - (x - \frac{x^3}{6}) = \frac{x^3}{2}$. We have $\tan(\tan x) - \tan x$. Let $y = \tan x$. As $x \to 0$, $y \to 0$. $\lim_{x \to 0} \frac{\tan(\tan x) - \tan x}{(\tan x)^3} = \frac{1}{3}$. So, $\tan(\tan x) - \tan x = \frac{1}{3}(\tan x)^3 = \frac{1}{3}x^3$.

We have $\sin x - \sin(\sin x)$. Let $z = \sin x$. As $x \to 0$, $z \to 0$. $\lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{(\sin x)^3} = -\frac{1}{6}$. So, $\sin(\sin x) - \sin x = -\frac{1}{6}(\sin x)^3 = -\frac{1}{6}x^3$.

Numerator: $\tan(\tan x) - \sin(\sin x) = (\tan(\tan x) - \tan x) + (\tan x - \sin x) + (\sin x - \sin(\sin x))$ $= \frac{1}{3}(\tan x)^3 + (\tan x - \sin x) - (\sin(\sin x) - \sin x)$ $= \frac{1}{3}x^3 + \frac{x^3}{2} - (-\frac{1}{6}x^3) = x^3(\frac{1}{3} + \frac{1}{2} + \frac{1}{6}) = x^3(\frac{2+3+1}{6}) = x^3(\frac{6}{6}) = x^3$.

Denominator: $\tan x - \sin x = \frac{x^3}{2}$.

So, the limit is $\lim_{x \to 0} \frac{x^3}{\frac{x^3}{2}} = 2$.

The original solution had a calculation error in the numerator's terms. The correct calculation gives 2.

Common Mistakes & Tips

• Incorrect Taylor Expansions: Ensure the correct order of Taylor series for $\sin x$ and $\tan x$ are used, up to the required order (usually $x^3$ for this type of problem).
• Algebraic Errors in Simplification: Carefully combine terms after substituting Taylor expansions. A small arithmetic mistake can lead to a wrong answer.
• Recognizing Standard Limits: Familiarity with standard limits like $\lim_{x \to 0} \frac{\tan x - x}{x^3} = \frac{1}{3}$ and $\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}$ can significantly speed up the calculation.

Summary

To find $f(0)$ for a continuous function, we need to evaluate the limit of $f(x)$ as $x$ approaches 0. The function $f(x)=\frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x}$ is in the indeterminate form $\frac{0}{0}$ at $x=0$. By using Taylor series expansions for $\tan x$ and $\sin x$ around $x=0$, we found that $\tan(\tan x) - \sin(\sin x)$ is approximately $x^3$ and $\tan x - \sin x$ is approximately $\frac{x^3}{2}$ for small $x$. The limit of the ratio of these approximations is $\frac{x^3}{x^3/2} = 2$. Therefore, $f(0) = 2$.

The final answer is \boxed{2}.