If the function f(x) = tan(tanx) − sin(sinx) / tanx − sinx is continuous at x = 0, then f(0) is equal to:

❓ 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 Solution

1. At $x=0$:
   $\tan(\tan 0) = \tan 0 = 0,\quad \sin(\sin 0)=\sin 0=0.$
   So numerator = $0-0=0$
   Denominator = $\tan 0 - \sin 0 = 0-0=0.$
   ⇒ Indeterminate form $0/0$. We must evaluate the limit as $x\to0$
   

2. Apply series expansion around $x=0$:

   • $\tan x = x + \tfrac{x^3}{3} + O(x^5).$
     

   • $\sin x = x - \tfrac{x^3}{6} + O(x^5).$
     

   Hence,
   $\tan x - \sin x = \Big(x+\tfrac{x^3}{3}\Big) - \Big(x-\tfrac{x^3}{6}\Big) + O(x^5) = \tfrac{x^3}{2} + O(x^5).$
   

3. Expand numerator:

   • $\tan(\tan x) \approx \tan\!\big(x+\tfrac{x^3}{3}\big).$
     Use $\tan u \approx u + u^3/3.$
     With $u=x+\tfrac{x^3}{3}$
     $\tan(\tan x) \approx (x+\tfrac{x^3}{3}) + \tfrac{1}{3}(x+\tfrac{x^3}{3})^3.$
     ≈ $x+\tfrac{x^3}{3} + \tfrac{x^3}{3} = x+\tfrac{2x^3}{3}.$
     

   • $\sin(\sin x) \approx \sin(x-\tfrac{x^3}{6}).$
     Use $\sin v \approx v - v^3/6.$
     With $v=x-\tfrac{x^3}{6}$
     $\sin(\sin x) \approx (x-\tfrac{x^3}{6}) - \tfrac{x^3}{6} = x-\tfrac{x^3}{3}.$
     

   So numerator = $\big(x+\tfrac{2x^3}{3}\big) - \big(x-\tfrac{x^3}{3}\big) = x+ \tfrac{2x^3}{3} -x + \tfrac{x^3}{3} = x^3.$
   

4. Now ratio:

   $$f(0)=\underset{x\to 0}{\lim }\frac{\text{numerator}}{\text{denominator}}=\underset{x\to 0}{\lim }\frac{x^3}{x^3\mathrm{/}2}=2.$$

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🖼️ Image Solution

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✅ Conclusion & Video Solution

Thus, by using Taylor expansions around zero, we resolve the indeterminate form and find:

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