If f(x)=limn→∞ ∑r=0n tan⁡x2r+1+tan3x2r+11−tan2⁡x2r+1 then limx→0 f(x)x is equal to - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
1
B
0
C
$- 1$
D
None of these

Solution:

Let $α_{r} = \frac{x}{2^{r + 1}}, r = 0, 1, 2, \dots, n .$ Then,

$\begin{array}{l} \frac{\tan \frac{x}{2^{r + 1}} + \tan^{3} \frac{x}{2^{r + 1}}}{1 - \tan^{2} \frac{x}{2^{r + 1}}} \\ = \frac{\tan α + \tan^{3} α_{r}}{1 - \tan^{2} α_{r}} \\ = \tan α_{r} (\frac{1 + \tan^{2} α_{r}}{1 - \tan^{2} α_{r}}) \end{array}$

$\begin{array}{l} = \frac{\tan α_{r}}{\cos 2 α_{r}} = \frac{\sin α_{r}}{\cos α_{r} \cos 2 α_{r}} \\ = \frac{\sin (2 α_{r} - α_{r})}{\cos α_{r} \cos 2 α_{r}} = \tan 2 α_{r} - \tan α_{r} \end{array}$

$∴ f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\tan 2 α_{r} - \tan α_{r})$

$\begin{array}{l} \Rightarrow f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\tan \frac{x}{2^{r}} - \tan \frac{x}{2^{r + 1}}) \\ \Rightarrow f (x) = lim_{n \to \infty} (\tan x - \tan \frac{x}{2^{n + 1}}) = \tan x \end{array}$

Hence, $lim_{x \to 0} \frac{f (x)}{x} = lim_{x \to 0} \frac{\tan x}{x} = 1$

If $f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} \frac{\tan \frac{x}{2^{r} + 1} + \tan^{3} \frac{x}{2^{r + 1}}}{1 - \tan^{2} \frac{x}{2^{r + 1}}}$ then $lim_{x \to 0} \frac{f (x)}{x}$ is equal to

Solution:

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