The integer n for which \large \lim_{x\rightarrow 0} \frac{(cos x -1)(cos x - e^{x})}{x^{n}}
is a finite non-zero number is
A. 1
B. 2
C. 3
D. 4
11Solve it by using expansion,
(cosx-1) (cosx-e x) x n
= (1/x n)[ (1- x2 / 2! + x4 / 4!- ...........) -1] [ (1- x2 / 2! + x 4 / 4! -....) - (1 + x + x2 / 2! + x3/3!......)]
=(1/x n) [ (-x2 / 2! + x4 / 4! - ........) (-x- x2 -x3/3! .......)]
= (-1 / x n-3) [ (-1 / 2! + x2 / 4!- .......) (1+x+x2/3!)...........)]
Therefore, for limit to exists we must have n-3 = 0 or n=3.
11any method to do this without expansion ,,i.e without expanding cos x and ex??
11I am not sure if I am correct or not,
we can do it by Hit and trial method. i mean by using L'h rule and side by side putting values given in (a), (b), (c), (d).
66We have
\dfrac{(\cos x-1)(\cos x-e^x)}{x^n} = \dfrac{(1-\cos x)(e^x-\cos x)}{x^n}
=\dfrac{(2\sin^2(x/2))(e^x-1+1-\cos x)}{x^n}
=\dfrac{1}{2}\dfrac{\sin^2(x/2)}{(x/2)^2}\cdot\left(\dfrac{e^x-1}{x}+\dfrac{2\sin^2(x/2)}{x}\right)\dfrac{1}{x^{n-3}}
For this limit to exist, n-3 ≤ 0 i.e. n ≤ 3.
But if n<3, the limit equals 0. So, for the limit to be non-zero, we must have n=3. In this case, the limit equals 1/2.
