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how to prove the right hand limit log(1 + 3x) /( e^2x - 1) is equal to 3/2 . . . pls help this is aa part of the limits question . . .just stuck with it. .pls help as soon as possible

how to prove the right hand limit    log(1 + 3x) /( e^2x - 1)   is equal to 3/2 . . . pls help this is aa part of the limits question  . . .just stuck with it.  .pls help as soon as possible

Grade:

3 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

jUst multiply and divide by 3x/2x

We know that log(1+x)/x=1

vikas askiitian expert
509 Points
10 years ago

L = log(1+3x)/e2x-1 lim x-> 0

for right hand limit  put x = 0+h

L = log(1+3(0+h))/e2(0+h)-1   lim h - >0

L = log1+3h/e2h-1   lim h->0

at h = 0 , this becomes 0/0 form...for solving such problems we have to differentiate nume &

denominator then put h->0 , this will give the limit

L = 3/(1+3h)(2(e2h))    lim h->0                (after differentiating)

on putting h=0 we get

L = 3/2

 

approve if u like this ans

rohit rathi
47 Points
10 years ago

You can use L-Hospital rule as it is 0/0 form:-

Differentiating numerator & denominator separately we get:-

3/2(1+3x).e^2x   ->then put x=0,you get ur ans.=3/2. If not understood read L-hospital rule from somewhere.

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