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Question Number 212745 by RoseAli last updated on 22/Oct/24

Answered by mehdee7396 last updated on 22/Oct/24

$${lim}_{x\to 0}(\frac{1+tanx}{1-tanx}-1)\times \frac{1}{sinx}$$$$={lim}_{x\to 0}\left(\frac{2tanx}{1-tanx}\right)\times \frac{1}{sinx}$$$$={lim}_{x\to 0}\left(\frac{2}{1-tanx}\right)\times \frac{1}{cosx}=2$$$$\Rightarrow ans=e^2✓$$$$=$$

Answered by depressiveshrek last updated on 22/Oct/24

$$\underset{x\to 0}{\lim }{(1+\frac{1+\tan x}{1-\tan x}-1)}^{\frac{1}{\sin x}}$$$$=\underset{x\to 0}{\lim }{(1+\frac{1+\tan x-(1-\tan x)}{1-\tan x})}^{\frac{1}{\sin x}}$$$$=\underset{x\to 0}{\lim }{\left({(1+\frac{2\tan x}{1-\tan x})}^{\frac{1-\tan x}{2\tan x}}\right)}^{\frac{2\tan x}{(1-\tan x)\sin x}}$$$$=e^{\underset{x\to 0}{\lim }\frac{2\tan x}{(1-\tan x)\sin x}\frac{:x}{:x}}$$$$=e^{2\cdot \underset{x\to 0}{\lim }\frac{\frac{\tan x}{x}}{(1-\tan x)\cdot \frac{\sin x}{x}}}$$$$=e^{2\cdot \frac{1}{(1-0)\cdot 1}}=e^2.$$

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