IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

madman

im getting the answer as 'e' but the answer is e/2

divalli_oct07

we can solve the problem using the series of log(1+x) and that of e^x..

let y=(1+x)^1/x

then, log y = 1/x log (1+x) (since log (1+x)=x - x^2/2 +x^3 /3...)

= 1/x[x - x^2/2+ x^3/3.......]

log y =1- x/2+ x^2/3................

therefore, y= e^1- x/2 + x^2/3 .....

y= e * e^ - x/2 + x^2/3....

y= e* [1+ (-x/2 +x^2/3) + (-x/2 +x^2/3)^2 /2] (since e^x=1+x+x^2/2!..)

y=e*[1-x/2 + 11x^2 /24]

Lt x tends to 0 [e(1-x/2 + 11x^2 /24) +ex - e] /x

Lt x tends to 0 [e - ex/2 + 11ex^2/24 + ex - e] /x

ans =e/2

madman

ya but suppose I did it this way
lim as x tends to zero of first part in the numerator is e.
then cancelling of the two e's i get ex in the numerator.
x cancels and i get e.

kulki_1123

i dont think u can just apply the limit in the numerator on one term and get the answer....if u do it then uve gotta do it with all the terms in that particular limit once the factor making the limit nonexistant dissapears....

so i agree with the cute baby one....

madman

just check this solution

bhartibhanushali

try by l hospital rule

divalli_oct07

@madman

u cannot do that way...

check out the last before step once again....

when u hav [e + _{[x ][ 0]}e(x - 1)]/x , how can u again apply limits to the

whole expression like this _{[x ][ 0]} e + e(x - 1)/x.

your procedure is totally wrong.

madman

why is it
limit as x tends to zero of e is 'e' itself so i can replace e with the limit
and then i am just replaceing the limits expression by a common limit
since it should again give the same thing
lim(f(x)+g(x)=limf(x)+lim(g(x)=lim(f(x)+g(x))
its the same thing

divalli_oct07

but in the solution u are doing it like this...pls observe....

_{[ ]}

_{[ ]} [F(x)+G(x)] =_{[ ]}

_{[ ]} F(x)+ _{[ ]}

_{[ ]}G(x) = _{[ ]}

_{[ ]}[value obtained after

applying the limits + G(x) ]

u are solving half part after applying the limits and again u are applying the limits to that solved part...see carefully...

panks_01

hey even i solved n got d lim of first part of num as 'e'

n then solving d second part i got it as 'ex'

nw since x tends to 0, ex also tends to 0

so even i got d ans as e..........

plz nudge me if u get d corr method n ans

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