IIT JEE , AIEEE Forum - Mathematics , Integral Calculus - Find this limit: a good exercise for all IIT aspirants

kaymant

Find this limit: a good exercise for all IIT aspirants

kaymant

hello guys.... anybody trying this one?

ankurgupta91

yup i m trying this one
actually i gt the answer bt its too long and i hope dat i m correct
i wl post the solution by 2mrw
so plz dont post it....

ankitagg

sorry that is 1/t

since it is 1^inf. form we use property

e^lim(t->0) {[a^t+1 - b^t+1]/(t+1)(a-b) - 1}*1/t

take LCM

now differentiating it ;since it is 0/0 form we get

e^ lim t->0 { a^(t+1).t.log a -b^(t+1).t.log b - a+b}/(2t+1)(a-b)

applying limit

e^ -(a-b)/(a-b)

e^-1

1/e ans.

sir please tell if any mistake is there

kaymant

well .... that's wrong actually.. but you tried well..

ankitagg

sir, where is my mistake .please tell me

ankitagg

ok sir i think i got my mistake . well.... will u please tell me the answer. (please dont post the whole solution)

kaymant

Re:Find this limit: a good exercise for all IIT aspirants

feynmann

Ok !! I got it !!

First plainly carry out the integration by putting y = (a-b)x + b

Then apply the formula (1+x ) ^(1/x) =e as x tends to zero .

ankurgupta91

i m nt writing limit again and again
so now its 1 raise to power infinity from
thus,
lim [ int(0to1) [1+(a-b)x+b]^t - 1 ]^(1/ t)

= e^ lim [ int(0 to 1)[[(a-b)x + b] ^t -1 ]/ t ]
nw diff it with respect to t

= e^ lim [ int(0 to 1) [(a-b)x+b]^t (log(a-b)x+b) ]
nw put t = 0
so
= e^ [ int(0 to 1) log(a-b)x+b)dx]
nw by integrating it as by applying by parts
and by putting the limits
we wl get the answer as

= (1/e)*[(a^a/ b^b)]^(1/ b-a)

sry in my pc equation editor was nt wrking
thats the answer.........

novelharry

hello could any one please the step for the given reply by ankur gupta after differentiating i could n't understand the 7 step

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