Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

We have,
$\underset{x\to 0}{\lim }\frac{(\cos x-1)\left(\cos x-e^x\right)}{x^n}$
$=\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right)\left(\frac{e^x-\cos x}{x^{n-2}}\right)$
$=\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right)\left(\frac{e^x-1+1-\cos x}{x}\right)\frac{1}{x^{n-3}}$
$=\underset{x\to 0}{\lim }\left(\frac{1-\cos x}{x^2}\right)\left\{\frac{e^x-1}{x}+\frac{1-\cos x}{x}\right\}\frac{1}{x^{n-3}}$
We know that
$\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}=\frac{1}{2},\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$
and $\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$
Therefore, for the above limit to be non-zero, we must have
$n-3=0$
$\Rightarrow n=3$