Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Evaluate the following limits limit x tends to 0 cos 2x - 1

17.   Evaluate the following limits  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

Answers (1)

best_answer

The limit:

  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit 

  \lim_{x\rightarrow 0} \frac{\cos 2x -1}{\cos x -1}

\lim_{x\rightarrow 0} \frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}

=\lim_{x\rightarrow 0}    \frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}

=\frac{1^2}{1^2}\times 4

= 4  (Answer)

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 supplementary admit card out; exam on July 28 for over 2.5 lakh
anam-khan

CBSE Class 12 supplementary exams 2026 datesheet out; test on July 28
anam-khan

Centre, CBSE, NCERT defend NEP three-language policy in SC; seek dismissal of pleas against framework

Latest Question