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Question Number 68368 by naka3546 last updated on 09/Sep/19

$$\frac{1}{x+1}=y$$$$\underset{x+1\to \mathrm{\infty }}{\lim }x\tan \left(\frac{1}{2x+2}\right)$$

Commented by kaivan.ahmadi last updated on 09/Sep/19

$$x+1\to \mathrm{\infty }\Rightarrow x\to \mathrm{\infty }$$$$and\frac{1}{2x+2}\to 0\Rightarrow tan\left(\frac{1}{2x+2}\right)\approx \frac{1}{2x+2}$$$$sowehave{lim}_{x\to \mathrm{\infty }}\frac{x}{2x+2}=\frac{1}{2}$$$$$$

Answered by Tanmay chaudhury last updated on 09/Sep/19

$$\underset{y\to 0}{\lim }(\frac{1}{y}-1)tan\left(\frac{y}{2}\right)$$$$\underset{y\to 0}{\lim }[\frac{1}{y}\times sin\left(\frac{y}{2}\right)\times \frac{1}{cos\left(\frac{y}{2}\right)}-tan\left(\frac{y}{2}\right)]$$$$\underset{y\to 0}{\lim }\frac{sin\left(\frac{y}{2}\right)}{\left(\frac{y}{2}\right)\times 2}\times \frac{1}{cos\left(\frac{y}{2}\right)}-\underset{y\to 0}{\lim }tan\left(\frac{y}{2}\right)$$$$=\frac{1}{2}\times \frac{1}{1}-0$$$$=\frac{1}{2}$$

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