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iit jee Grade: 12
```        evaluate without using series
lim     x tan 2x - 2x tan x
x -0    (1 - cos 2x )```
7 years ago

## Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
```										Dear
Lt x→o {xtan2x -2xtanx}/{1-cos2x}
= Lt x→o {xtan2x  -2xtanx}/{2sin2x}
=Lt x→o {(tan2x   -2tanx)/2x}/{(sinx)/x}2
=Lt  x→o {(tan2x   -2tanx)/2x}           Lt  x→o sinx /x =1
= Lt  x→o tan2x /2x      -2tanx/2x
= 1 -1  =0
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get youthe answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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