Find the integer 'n' for which the the

 is finite non-zero number.

use L'Hosp:

it is initially 0/0 form...so diff. u get some a/b

now the numerator @ x=0 will be = -1...so denom shud be finite...denom is n*x^(n-1)..for this to be finite @ x=0,n-1 shud be=0 so n=1

n can be 2 also

ANSWER=4

yes i m also getting the ans as  4 but i wasnt sure but now i m

see we will apply the sereis of e^x and cos x  so taking the 1st approximation we will get

e^x=1+x and cos x=1-x²/2 and put in the eqn so taking the highest degree term we get ans as 4

hope u got it....

Could someone pls work this out fully for me and prove conclusively that n can be 4.

it is not in the form of of 0/0.so you cannot use L-HOSPITAL.if i am wrong please correct me.

yes friend it is of the form 0/0  if put  x=0

plz check carefully

Yes , questions is right with the answer that i have posted

0/0 form ..so use L'Hosp...if u diff once or twiceor thrice u stlill get a 0/0 form

on 4th diff u get:

{8cos2x  -cosx + 4cosx*e^x + e^x}/n*n-1*n-2*n-3*x^n-4

numerator is non 0 hence denom shud be non 0 hence n-4=0 so n=4

@ bhattji n=2 doesn work...

@ akku...for n=0 the lim is 0...but the q states tht the lim shud be a non-zero value