Question

The positive integer $n$ for which $\displaystyle\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^{x}\right)}{x^{n}}$ exists and is finite, is

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

Let
$\displaystyle\lim _{x \rightarrow 0} \frac{(\cos x-1)\left(\cos x-e^{x}\right)}{x^{n}}=K$
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{(1-\cos x)\left(e^{x}-1-\cos x+1\right)}{x^{n}}=K$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{n-1}}\left(\frac{e^{x}-1}{x}+\frac{1-\cos x}{x}\right)=K$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0} \frac{2 \sin ^{2} \frac{x}{2}}{x^{n-1}}\left(\frac{e^{x}-1}{x}+\frac{2 \sin ^{2} \frac{x}{2}}{x}\right)=K$
$\Rightarrow \displaystyle \lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\left(\frac{2 \sin ^{2} \frac{x}{2}}{x^{n-1}}\right)+\displaystyle\lim _{x \rightarrow 0} \frac{4 \sin ^{4} \frac{x}{2}}{x^{n}}=K$
$\Rightarrow \displaystyle\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} \frac{x}{2}}{x^{n-1}}\right)+\lim _{x \rightarrow 0} \frac{4 \sin ^{2} \frac{x}{2}}{x^{n}}=K$
$\left[\because \displaystyle\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right]$
$=\displaystyle\lim _{x \rightarrow 0} \frac{1}{2\left(x^{n-3}\right)}+\displaystyle\lim _{x \rightarrow 0} \frac{1}{4\left(x^{n-4}\right)}=K$
To exists and finite put $n-3=0 $
$\Rightarrow n=3$