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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
Dear student,
jUst multiply and divide by 3x/2x
We know that log(1+x)/x=1
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
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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