lim (e^x - x)
---------
1 - cosx
x-->0
(Edited by INDIANET @IIT-JEE - Thursday, 13 April 2006, 10:21 AM)
Please frame your question properly.
Well, teja krishna has framed the ques. correctly
u may see his ques.
u may see his ques.
if your question is Lt x------>0 ex-x/1-cosx
then by direct application of limit we get 1-0/1-1
i.e. 1/0 i.e. infinite.
you can't apply L'Hospital's rule here because it is applicable only to indeterminate forms and that too 0/0
and infinity/infinity forms. hope i dont confuse you. but i think infinity is the answer.
sry teja krishna the answer is 1. Well as far as i know infinity is the answer when limit gives answer undefined but not indeterminate. The answer will come only by applying L'Hospital's rule only. But i don't know how.
Solution given by teja krishna is cent percent correct.
I'm very much sorry. Well the correct ques. is: limitx--->0 log(ex-x)/(1-cosx)
Very sorry to write the ques. wrongly. Plz solve now, the answer is 1.
Very sorry to write the ques. wrongly. Plz solve now, the answer is 1.
Directly substitute x=0
we get answer 0.So may be given answer is wrong.
Shashank dear, if u substitute x=0 in log(ex-x)/(1-cosx) then we get ->0/->0 which is indeterminate. Plz try it again. Its a correct ques. with answer 1 only. Try doing it by L'Hospital's rule.
(Edited by INDIANET @IIT-JEE - Tuesday, 18 April 2006, 09:57 PM)
use l-hospital
[{1/(e^x-x)}(e^x-1)]/sin x
this is again 0/0 form so use l-hospital again
u/v method
u get-:
[{(e^x-x)e^x-(e^x-1)^2}/(e^x-x)^2]/cos x
now putting x=0 u get ans 1,i hope i m right.
Shaan, You are right. Thanks for your solution. Can you try doing it without L-hospital's rule but with help of ex expansion. One of my friend did it without L-Hospital's rule. Isn't it amazing. Please try it.
Hey Shaan,
I got the method for doing this ques. without L-Hospital's rule.
[log(ex-x)]/(1-cosx) = [log(1+(ex-x -1)] x (ex-x -1) / [(ex-x -1)(1-cosx)]
now, lim log(1+(ex-x -1))/(ex-x -1) = 1
x-->0
Now, expanding (ex-x -1)/(1-cosx) we get (1 + x/1! + x2/2! + x3/3! +... - x - 1)/(1 - (1 - x2/2! + x4/4! - ....))
= (x2/2! + x3/3! +..)/(x2/2! - x4/4! + ....)
= (1/2! + x/3! + ...)/(1/2! - x2/4! + ...) = h(x) say
lim h(x) = (1/2)/(1/2) = 1
x-->0
Therefore, lim of f(x) for x-->0 is 1 x 1 = 1 Ans.
I got the method for doing this ques. without L-Hospital's rule.
[log(ex-x)]/(1-cosx) = [log(1+(ex-x -1)] x (ex-x -1) / [(ex-x -1)(1-cosx)]
now, lim log(1+(ex-x -1))/(ex-x -1) = 1
x-->0
Now, expanding (ex-x -1)/(1-cosx) we get (1 + x/1! + x2/2! + x3/3! +... - x - 1)/(1 - (1 - x2/2! + x4/4! - ....))
= (x2/2! + x3/3! +..)/(x2/2! - x4/4! + ....)
= (1/2! + x/3! + ...)/(1/2! - x2/4! + ...) = h(x) say
lim h(x) = (1/2)/(1/2) = 1
x-->0
Therefore, lim of f(x) for x-->0 is 1 x 1 = 1 Ans.
I am sorry.
But even then L'Hopital is valid only for 0/0 or infinity/infinity form.Here I see that numerator is not 0 on subs.x=0.It is (e^0)-0=1
But log(1) is 0. it was log(ex-x) and not just (ex-x). Now try solving it. Well Shaan has already given me the soln.