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Question Number 59599 by tanmay last updated on 12/May/19

Answered by tanmay last updated on 12/May/19

$$T_r=\frac{tan\frac{x}{2^{r+1}}(1+{tan}^2\frac{x}{2^{r+1}})}{1-{tan}^2\frac{x}{2^{r+1}}}$$$$T_r=\frac{tan\frac{x}{2^{r+1}}}{cos2\times \frac{x}{2^{r+1}}}$$$$=\frac{tan\frac{x}{2^{r+1}}}{cos\frac{x}{2^r}}$$$$f\left(x\right)=\underset{n\to \mathrm{\infty }}{\lim }(\frac{tan\frac{x}{2}}{cos\frac{x}{2^0}}+\frac{tan\frac{x}{2^2}}{cos\frac{x}{2}}+\frac{tan\frac{x}{2^3}}{cos\frac{x}{2^2}}+..+\frac{tan\frac{x}{2^{n+1}}}{cos\frac{x}{2^n}})$$$$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}$$$$\underset{n\to \mathrm{\infty }}{\lim }(\frac{tan\frac{x}{2}}{\frac{x}{2}\times 2\times cos\frac{x}{2^0}}+\frac{tan\frac{x}{2^2}}{\frac{x}{2^2}\times 2^2\times cos\frac{x}{2}}+...+\frac{tan\frac{x}{2^{n+1}}}{\frac{x}{2^{n+1}}\times 2^{n+1}\times cos\frac{x}{2^n}})$$$$=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..\mathrm{\infty }\left(gpseries\right)$$$$=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$$

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