Animated Solution for Mathematics - Limits, Continuity and Differentiability: 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 ________

A Journey into Continuity

Welcome, students! Today, we are going to peel back the layers of a truly fascinating calculus problem. We are presented with a function

and told it is continuous at $x=0$.

Our goal is to find $f(0)$. Remember, for a function to be continuous at a point, the value of the function at that point must equal the limit of the function as it approaches that point. So, our mission is to evaluate ${\lim }_{x\to 0}f(x)$.

The Foundation (The Denominator)

Before we tackle the intimidating numerator, let's look at the denominator: $D=\tan x-\sin x$. When $x$ is near zero, we can use the Maclaurin series expansions:

Substituting these into our denominator, we get $D\approx (x+\frac{x^3}{3})-(x-\frac{x^3}{6})$. Notice how the linear $x$ terms cancel out beautifully!

We are left with $\frac{x^3}{3}+\frac{x^3}{6}=\frac{x^3}{2}$. This is a powerful, standard approximation: near zero, $\tan x-\sin x$ behaves exactly like $\frac{x^3}{2}$.

The Divide and Conquer Strategy

Now, consider the numerator: $\tan (\tan x)-\sin (\sin x)$. Expanding this directly would be a nightmare.

Instead, we use a clever algebraic manipulation. We add and subtract $\tan x$ and $\sin x$ to split the limit into three distinct, manageable parts: $L_1+L_2+L_3$, where:

Solving the Pieces

For $L_1$, we use the standard limit ${\lim }_{u\to 0}\frac{\tan u-u}{u^3}=\frac{1}{3}$. By multiplying and dividing by $(\tan x{)}^3$, we isolate this form.

Since $\tan x\approx x$ and the denominator $\tan x-\sin x\approx \frac{x^3}{2}$, $L_1$ becomes:

For $L_2$, it is a simple cancellation: $\frac{\tan x-\sin x}{\tan x-\sin x}=1$. This is the easiest piece of the puzzle!

For $L_3$, we use the standard limit ${\lim }_{u\to 0}\frac{u-\sin u}{u^3}=\frac{1}{6}$. Again, multiplying and dividing by $(\sin x{)}^3$, we find:

Finally, we sum our results:

Combining the fractions, $\frac{2}{3}+\frac{1}{3}=1$. Thus, $f(0)=1+1=2$.

We have conquered the monster! By breaking it down into smaller, standard forms, we turned an intimidating expression into a simple, elegant solution. Keep this strategy in your toolkit—it is the secret to mastering JEE calculus.