Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Evaluating <img alt="\mathrm{\lim _{x \rightarrow 0} \frac{\ln (1+x)-\ln (1-x)}{\arctan (1+x)-\arctan (1-x)}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5Cr

Evaluating \mathrm{\lim _{x \rightarrow 0} \frac{\ln (1+x)-\ln (1-x)}{\arctan (1+x)-\arctan (1-x)}}

Option: 1

0


Option: 2

3


Option: 3

2


Option: 4

1


Answers (1)

best_answer

\mathrm{\mathrm{m}_{\rightarrow 0} \frac{\ln (1+x)-\ln (1-x)}{\arctan (1+x)-\arctan (1-x)} =\lim _{x \rightarrow 0} \frac{\ln \left(1+\frac{2 x}{1-x}\right)}{\arctan \left(\frac{2 x}{2-x^{2}}\right)} }
                                                                            \mathrm{=\lim _{x \rightarrow 0}\left(\frac{\ln \left(1+\frac{2 x}{1-x}\right)}{\frac{2 x}{1-x}}\right) \lim _{x \rightarrow 0}\left(\frac{\frac{2 x}{2-x^{2}}}{\arctan \left(\frac{2 x}{2-x^{2}}\right)}\right) \lim _{x \rightarrow 0} \frac{\frac{2 x}{1-x}}{\frac{2 x}{2-x^{2}}} }
                                                                          \mathrm{=\text { 1.1. } \lim _{x \rightarrow 0} \frac{2-x^{2}}{1-x}}
                                                                          \mathrm{=2}

Posted by

manish

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question