Question

If $\alpha =\displaystyle\lim_{x\to0} \frac{x.2^{x}-x}{1-\cos x} $ and $ \beta =\displaystyle\lim_{x\to0} \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,
$\alpha=\displaystyle\lim _{x \rightarrow 0} \frac{x \cdot 2^{x}-x}{1-\cos x}$
Applying L'Hospital rule
$\alpha=\displaystyle\lim _{x \rightarrow 0} \frac{2^{x}+x \cdot 2^{x} \log 2-1}{\sin x}$
Again, applying L'Hospital rule
$\alpha=\displaystyle\lim _{x \rightarrow 0} \frac{2^{x} \log 2+2^{x} \log 2+x \cdot 2^{x}(\log 2)^{2}}{\cos x}$
$\alpha=\log 2+\log 2 \alpha=2 \log 2 \dots$(i)
and $\beta =\displaystyle\lim _{x \rightarrow 0} \frac{x \cdot 2^{x}-x}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}$
$=\displaystyle\lim _{x \rightarrow 0} \frac{2^{x}+x \cdot 2^{x} \log 2-1}{\left(\frac{2 x}{2 \sqrt{1+x^{2}}}+\frac{2 x}{2 \sqrt{1-x^{2}}}\right)} $
(By L'Hospital rule)
Again, applying L'Hospital rule, we get
$=\displaystyle\lim _{x \rightarrow 0} \frac{2^{x} \log 2+2^{x} \log 2+x \cdot 2^{x}(\log 2)^{2}}{ \left.\sqrt{1+x^{2}}-\left(\frac{2 x^{2}}{2 \sqrt{1+x^{2}}}\right)\right] +\left[ \sqrt{1 - x^2} + \left(\frac{2x^2}{2\sqrt{1 - x^2}}\right)\right]}$
$= \frac{2 \log 2}{(1-0)+(1+0)}=\frac{2 \log 2}{2}$
$\beta=\log 2 \dots$(ii)
By Eqs. (i) and (ii), we get
$\alpha=2 \beta$