Question

The integer n for which $\underset{x\to 0}{lim}\frac{(\cos x-1)(\cos x-e^x)}{x^n}$ is a finite nonzero is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

$\frac{(\cos x-1)(\cos x-e^x)}{x^n}$
$=\frac{1}{x^n}((1-\frac{x^2}{2!}+\frac{x^4}{4!}-...)-1)((1-\frac{x^2}{2!}+\frac{x^4}{4!}-...)-(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}...))$
$=\frac{1}{x^n}\left[(-\frac{x^2}{2!}+\frac{x^4}{4!}-....)(-x-x-\frac{x^3}{3!}....)\right]$
$=\frac{-1}{x^{n-3}}\left[(-\frac{1}{2!}+\frac{x^2}{4!}-...)(1+x+\frac{x^2}{3!})\right]$
$\therefore \underset{x\to 0}{lim}\frac{(\cos x-1)(\cos x-e^x)}{x^n}$ to exist we must have $n-3=0\Rightarrow n=3$
Hence, option 'C' is correct.