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Question Number 55149 by pooja24 last updated on 18/Feb/19

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19

$$\underset{x\to 0}{\mathrm{li}}$$$$x=tan\theta asx\to 0\theta \to 0$$$$\underset{\theta \to 0}{\lim }\frac{1}{ln(tan\theta +sec\theta )}-\frac{1}{ln(1-tan\theta )}$$$$\underset{\theta \to 0}{\lim }\frac{1}{\left\{ln\left(\frac{1+sin\theta }{cos\theta }\right)\right\}}-\frac{1}{ln(1-tan\theta )}$$$$t=tan\frac{\theta }{2}as\theta \to 0t\to 0$$$$\underset{t\to 0}{\lim }\frac{1}{\left\{ln\left(\frac{1+\frac{2t}{1+t^2}}{\frac{(1+t)(1-t)}{1+t^2}}\right)\right\}}-\frac{1}{ln(1-\frac{2t}{1-t^2})}$$$$\underset{t\to 0}{\lim }\frac{1}{ln\left(\frac{1+t}{1-t}\right)}-\frac{1}{ln(1-\frac{2t}{1-t^2})}$$$$\underset{t\to 0}{\lim }\frac{1}{p}-\frac{1}{q}$$$$now..\underset{t\to 0}{\lim }\frac{1}{p}valueis$$$$\underset{t\to 0}{\lim }\frac{1}{\frac{ln(1+t)}{t}\times t-\frac{ln(1-t)}{-t}\times -t}$$$$\underset{t\to 0}{\lim }\frac{1}{t+t}\to \underset{t\to 0}{\lim }\frac{1}{2t}....\left[wait\right]$$$$now\underset{t\to 0}{\lim }\frac{1}{q}$$$$\underset{t\to 0}{\lim }\frac{1}{ln(1-\frac{2t}{1-t^2})}$$$$\underset{t\to 0}{\lim }\frac{1}{\left[\frac{ln(1-\frac{2t}{1-t^2})}{\frac{-2t}{1-t^2}}\right]\times \frac{-2t}{1-t^2}}$$$$\underset{t\to 0}{\lim }\frac{1-t^2}{-2t}$$$$nowcombining...$$$$\underset{t\to 0}{\lim }[\frac{1}{2t}-\left(\frac{1-t^2}{-2t}\right)]$$$$\underset{t\to 0}{\lim }\frac{1+1-t^2}{2t}$$$$usingLHrule$$$$\underset{t\to 0}{\lim }\frac{0-2t}{2}$$$$soansweris0$$$$plsothercheck...$$$$$$$$$$$$$$

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