Question

Let $f\left(x\right)=\frac{x.2^x-x}{1-cosx}$ & $g\left(x\right)=2^x$ sin$\left(\frac{ln2}{2^x}\right)$ then

• A. $\underset{x\to 0}{lim}f\left(x\right)=ln2$
• B. $\underset{x\to 0}{lim}g\left(x\right)=ln4$
• C. $\underset{x\to 0}{lim}f\left(x\right)=ln4$
• D. $\underset{x\to 0}{lim}g\left(x\right)=ln2$

Answer 1

Answer: (C), (D)

$\underset{x\to 0}{lim}g\left(x\right)=ln2$
$\underset{x\to 0}{lim}f\left(x\right)=ln4$

$f\left(x\right)=\frac{2^x-1}{\frac{2si{n}^2\frac{x}{2}}{x}}$
$=\frac{2^x-1}{\frac{x}{4}.\frac{2si{n}^2\frac{x}{2}}{\frac{x^2}{4}}}$
$=\frac{2^x-1}{\frac{x}{2}.\frac{si{n}^2\frac{x}{2}}{\frac{x^2}{4}}}$

As $x\to 0$

$f\left(x\right)=2ln\left(2\right)$
$=ln\left(4\right)$.

$\underset{x\to 0}{lim}f\left(x\right)=ln4$

Now
$g\left(x\right)=\frac{sin\left(\frac{ln\left(2\right)}{2^x}\right)}{\frac{ln\left(2\right)}{2^x}}.ln\left(2\right)$

As $x\to 0$
$g\left(x\right)=ln\left(2\right)$.

$\underset{x\to 0}{lim}g\left(x\right)=ln2$