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```				   the value of lim x  tands to  infifnite of
(e^x+x)^1/x is
```

6 years ago

Share

```										Dear manan
y =Lt x→∞ (ex+x)1/x
or
lny = Lt x→∞ ln[(ex+x)]/x
=Lt x→∞ ln[(ex+x)]/x
use L hospital rule
lny =Lt x→∞ (1+ex)/(ex+x)
again
lny = Lt x→∞ ex/(ex+1)
or
lny =Lt x→∞ 1/(e-x+1)
apply limit
lny =1
y =e

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
```
6 years ago

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• Complete JEE Main/Advanced Course and Test Series
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