Limits, Continuity & Differentiability Question - None

Question

If $$f(x)$$ is odd linear polynomial with $$f(1) = 1$$ , then $\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{x^2 f(\sin x)}$ is

Accepted Answer

Key Idea: First determine $$f(1) = 1$$ , then use $\lim_{x \to 0} \frac{2^{f(\tan x)} - 2^{f(\sin x)}}{x^2 f(\sin x)}$ and $an x - \sin x \approx x^3/2$ .

Step 1: Find $$f(x)$$ . Odd linear polynomial means $$f(-x) = -f(x)$$ , so $$f(x) = cx$$ . With $$f(1) = 1$$ : $$f(x) = x$$ .

Step 2: Substitute $$f(x) = x$$ :
$\lim_{x o 0}\frac{2^{ an x} - 2^{\sin x}}{x^2 \sin x}$

Step 3: Factor out $2^{\sin x}$ :
$= \lim_{x o 0}\frac{2^{\sin x}\left(2^{ an x - \sin x} - 1 ight)}{x^2 \sin x}$ As $$x o 0$$ : $2^{\sin x} o 1$ and $\sin x \sim x$ .

Step 4: Use $2^u - 1 \sim u\ln 2$ as $$u o 0$$ :
$= \lim_{x o 0}\frac{( an x - \sin x)\ln 2}{x^3}$

Step 5: Use $an x - \sin x = \frac{\sin x(1-\cos x)}{\cos x} \approx \frac{x \cdot x^2/2}{1} = \frac{x^3}{2}$ :
$= \frac{(x^3/2)\ln 2}{x^3} = \frac{\ln 2}{2}$

Answer: (C) $\dfrac{1}{2}\ln 2$

Trap: Two traps here — (1) not identifying

$f(x) = x$

from the odd linear condition, and (2) forgetting that

an x - \sin x \sim x^3/2

, not 0. Both cause wrong answers.

Key Concept: Odd function identification + standard limit

2^u - 1 \sim u\ln 2

+

an x - \sin x

expansion

Question:

Step-by-Step Solution

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