Limits, Continuity & Differentiability
Telescoping Limit / Standard Limit
nta_pyq_2025_apr
Grade 12
Question:
Let $f(x) = \lim_{n \to \infty} \displaystyle\sum_{r=0}^{n}\left(\frac{\tan(x/2^{r+1}) + \tan^3(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})}\right)$. Then $\lim_{x \to 0} \dfrac{e^x - e^{f(x)}}{x - f(x)}$ is equal to:
Step-by-Step Solution
Key Concept: Recognize that $\frac{\tan\theta+\tan^3\theta}{1-\tan^2\theta} = \frac{\tan\theta(1+\tan^2\theta)}{1-\tan^2\theta} = \tan2\theta - \tan\theta$ (using double angle). The sum telescopes to $\tan x$.
Each term $= \tan(x/2^r)-\tan(x/2^{r+1})$ (telescoping). Sum $= \tan x - \tan(x/2^{n+1}) \to \tan x$ as $n\to\infty$. So $f(x)=\tan x$. $\lim_{x\to0}\frac{e^x-e^{\tan x}}{x-\tan x} = \lim_{x\to0}e^{\tan x}\cdot\frac{e^{x-\tan x}-1}{x-\tan x} = 1\cdot1 = 1$.
Correct Answer: 1