Let f(x)= _ n arrow ∞ ∑_ r=0 ^ n ( (x / 2^ r+1 )+ ^3(x / 2^ r+1 ) 1- ^2(x / 2^ r+1 ) ) . Then _ x arrow 0 e^x-e^ f(x) (x-f(x)) is equal to

Correct answer: 0 aligned & f(x)= _ x arrow ∞ ( (x / 2^ r+1 )+ ^3(x / 2^ r+1 ) 1- ^2(x / 2^ r+1 ) ) & = _ x arrow ∞ ( x 2^ r+1 ) ((x)/(2^r)) & = _ x arrow ∞ (x / 2^ r+1 ) ( x 2^ r+1 ) ((x)/(2^r)) & = _ x arrow ∞ ((x)/(2^r)- x 2^ r+1 ) ( x 2^ r+1 ) ((x)/(2^r)) aligned = _ x arrow ∞ ((x)/(2^r))- ( x 2^ r+1 ) From condition given question aligned & _ n arrow ∞ ∑_ r=0 ^n[ ((x)/(2^r))- ( x 2^ r+1 )]= x & _ x arrow 0 ( e^x-e^ x x- x ) & _ x arrow 0 e^ x ( e^ x- x -1 x- x ) & _ x arrow 0 e^ x _ x arrow 0 ( e^ x- x -1 x- x ) & 1.1( _ x arrow 0 e^ x-1 x =1) & =1 aligned

JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2025 · 28 January · Evening Shift · Mathematics · Limits

Let f(x)= _ n arrow ∞ ∑_ r=0 ^ n ( (x / 2^ r+1 )+ ^3(x / 2^ r+1 )...

Let

f(x)=\lim _{\mathrm{n} \rightarrow \infty} \sum_{\mathrm{r}=0}^{\mathrm{n}}\left(\frac{\tan \left(x / 2^{r+1}\right)+\tan ^3\left(x / 2^{r+1}\right)}{1-\tan ^2\left(x / 2^{r+1}\right)}\right)

. Then

\lim _{x \rightarrow 0} \frac{\mathrm{e}^x-\mathrm{e}^{f(x)}}{(x-f(x))}

is equal to

Correct answer

Step-by-step solution

\begin{aligned} & f(x)=\lim _{x \rightarrow \infty}\left(\frac{\tan \left(x / 2^{r+1}\right)+\tan ^3\left(x / 2^{r+1}\right)}{1-\tan ^2\left(x / 2^{r+1}\right)}\right) \\ & =\lim _{x \rightarrow \infty} \frac{\tan \left(\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^r}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\sin \left(x / 2^{r+1}\right)}{\cos \left(\frac{x}{2^{r+1}}\right) \cos \left(\frac{x}{2^r}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{x}{2^r}-\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^{r+1}}\right) \cos \left(\frac{x}{2^r}\right)}\end{aligned}

=\lim _{x \rightarrow \infty} \tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)

From condition given question

\begin{aligned} & \therefore \lim _{n \rightarrow \infty} \sum_{r=0}^n\left[\tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)\right]=\tan x \\ & \therefore \lim _{x \rightarrow 0}\left(\frac{e^x-e^{\tan x}}{x-\tan x}\right) \\ & \Rightarrow \lim _{x \rightarrow 0} e^{\tan x}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \\ & \Rightarrow \lim _{x \rightarrow 0} e^{\tan x} \lim _{x \rightarrow 0}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \\ & \Rightarrow 1.1\left(\because \lim _{x \rightarrow 0} \frac{e^{x-1}}{x}=1\right) \\ & \quad=1 \end{aligned}

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