Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (B)

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