Question

The integer n for which $\displaystyle \lim_{x \to0} \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,
$\displaystyle\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^{x}\right)}{x^{n}}$
$=\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)\left(\frac{e^{x}-\cos x}{x^{n-2}}\right)$
$=\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)\left(\frac{e^{x}-1+1-\cos x}{x}\right) \frac{1}{x^{n-3}}$
$=\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^{2}}\right)\left\{\frac{e^{x}-1}{x}+\frac{1-\cos x}{x}\right\} \frac{1}{x^{n-3}}$
We know that
$\displaystyle\lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}=\frac{1}{2}, \displaystyle\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1 $
and $ \displaystyle\lim _{x \rightarrow 0} \frac{1-\cos x}{x}=0$
Therefore, for the above limit to be non-zero, we must have
$n -3=0 $
$\Rightarrow n =3$