Continuity at Zero for Tan and Sin Expression

❓ Question:

If the function

$$f(x)=\frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin \mathrm{x}}$$

is continuous at $x=0$, then find the value of $f(0)$.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Continuity of functions
👉 Standard limits
👉 Expansion method.

Main idea:

For continuity at:

$$x=0$$

$$\boxed{ f(0)=\lim_{x\to0}f(x) }$$

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🔷 Step 1 — Write Required Limit 💯

We need:

$$f(0)= \lim_{x\to0} \frac{\tan(\tan x)-\sin(\sin x)} {\tan x-\sin x}$$

As:

$$x\to0$$

both numerator and denominator become:

$$0$$

So it is:

$$\frac00$$

form.

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🔷 Step 2 — Use Small Angle Expansions

Recall:

$$\tan y=y+\frac{y^3}{3}+...$$
$$\sin y=y-\frac{y^3}{6}+...$$

Let:

$$a=\tan x$$
$$b=\sin x$$

Then:

$$\tan(\tan x)=a+\frac{a^3}{3}$$
$$\sin(\sin x)=b-\frac{b^3}{6}$$

Thus numerator:

$$= (a-b)+\frac{a^3}{3}+\frac{b^3}{6}$$

So:

$$f(x) = \frac{ (a-b)+\frac{a^3}{3}+\frac{b^3}{6} } {a-b}$$
$$= 1+ \frac{ \frac{a^3}{3}+\frac{b^3}{6} } {a-b}$$

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🔷 Step 3 — Approximate $a-b$

Using standard expansions:

$$\tan x=x+\frac{x^3}{3}$$
$$\sin x=x-\frac{x^3}{6}$$

Therefore:

$$a-b = \frac{x^3}{3}+\frac{x^3}{6}$$
$$= \frac{x^3}{2}$$

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🔷 Step 4 — Simplify Numerator Correction Term

Also:

$$a^3\approx x^3$$
$$b^3\approx x^3$$

Thus:

$$\frac{a^3}{3}+\frac{b^3}{6} = \frac{x^3}{3}+\frac{x^3}{6}$$
$$= \frac{x^3}{2}$$

Hence:

$$f(x) = 1+ \frac{x^3/2}{x^3/2}$$
$$=1+1$$
$$=2$$

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🔷 Step 5 — Continuity Condition

For continuity:

$$\boxed{ f(0)=\lim_{x\to0}f(x) }$$

Thus:

$$\boxed{ f(0)=2 }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Direct substitution kar dena

❌ Small angle expansion galat use karna

❌ Higher order terms unnecessarily retain kar lena

Remember:

$$\boxed{ \tan x=x+\frac{x^3}{3} }$$
$$\boxed{ \sin x=x-\frac{x^3}{6} }$$

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✅ Final Answer

$$\boxed{{\displaystyle 2}}$$

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