Question

If $\alpha =\underset{x\to 0}{\lim }\frac{x.2^x-x}{1-\cos x}$ and $\beta =\underset{x\to 0}{\lim }\frac{x.2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}},$ then

• A. $\alpha =5\beta$
• B. $\alpha =2\beta$
• C. $\beta =2{\alpha }^2$
• D. $\beta =\frac{1}{6}\alpha$

Answer 1

Answer: (B)

Given,
$\alpha =\underset{x\to 0}{\lim }\frac{x\cdot 2^x-x}{1-\cos x}$
Applying L'Hospital rule
$\alpha =\underset{x\to 0}{\lim }\frac{2^x+x\cdot 2^x\log 2-1}{\sin x}$
Again, applying L'Hospital rule
$\alpha =\underset{x\to 0}{\lim }\frac{2^x\log 2+2^x\log 2+x\cdot 2^x(\log 2{)}^2}{\cos x}$
$\alpha =\log 2+\log 2\alpha =2\log 2\dots$(i)
and $\beta =\underset{x\to 0}{\lim }\frac{x\cdot 2^x-x}{\sqrt{1+x^2}-\sqrt{1-x^2}}$
$=\underset{x\to 0}{\lim }\frac{2^x+x\cdot 2^x\log 2-1}{\left(\frac{2x}{2\sqrt{1+x^2}}+\frac{2x}{2\sqrt{1-x^2}}\right)}$
(By L'Hospital rule)
Again, applying L'Hospital rule, we get
$=\underset{x\to 0}{\lim }\frac{2^x\log 2+2^x\log 2+x\cdot 2^x(\log 2{)}^2}{\sqrt{1+x^2}-\left(\frac{2x^2}{2\sqrt{1+x^2}}\right)]+\left[\sqrt{1-x^2}+\left(\frac{2x^2}{2\sqrt{1-x^2}}\right)\right]}$
$=\frac{2\log 2}{(1-0)+(1+0)}=\frac{2\log 2}{2}$
$\beta =\log 2\dots$(ii)
By Eqs. (i) and (ii), we get
$\alpha =2\beta$