If f(x)=limn→∞ ∑r=0n tan⁡x2r+1+tan3x2r+11−tan2⁡x2r+1 then limx→0 f(x)x is equal to

# If $f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim} \sum _{r=0}^{n} \frac{\mathrm{tan}\frac{x}{{2}^{r}+1}+{\mathrm{tan}}^{3}\frac{x}{{2}^{r+1}}}{1-{\mathrm{tan}}^{2}\frac{x}{{2}^{r+1}}}$ then $\underset{x\to 0}{lim} \frac{f\left(x\right)}{x}$ is equal to

1. A

1

2. B

0

3. C

$-1$

4. D

None of these

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### Solution:

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

$\begin{array}{l}=\frac{\mathrm{tan}{\alpha }_{r}}{\mathrm{cos}2{\alpha }_{r}}=\frac{\mathrm{sin}{\alpha }_{r}}{\mathrm{cos}{\alpha }_{r}\mathrm{cos}2{\alpha }_{r}}\\ =\frac{\mathrm{sin}\left(2{\alpha }_{r}-{\alpha }_{r}\right)}{\mathrm{cos}{\alpha }_{r}\mathrm{cos}2{\alpha }_{r}}=\mathrm{tan}2{\alpha }_{r}-\mathrm{tan}{\alpha }_{r}\end{array}$

Hence, $\underset{x\to 0}{lim} \frac{f\left(x\right)}{x}=\underset{x\to 0}{lim} \frac{\mathrm{tan}x}{x}=1$

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