Question

$\underset{x\to 0}{lim}\frac{1-\cos \left(x^2\right)}{x^3(4^x-1)}$ is equal to

• A. $\frac{1}{4\log 2}$
• B. $\frac{1}{2\log 4}$
• C. $\log 4$
• D. $1-\frac{1}{2}\log \left(\frac{e^2}{4}\right)$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{4\log 2}$
B $\frac{1}{2\log 4}$

$\underset{x\to 0}{lim}\frac{1-\cos \left(x^2\right)}{x^3(4^x-1)}$
$={lim}_{x\to 0}\frac{2{\sin }^2\left(\frac{x^2}{2}\right)}{x^3(4^x-1)}$
$={lim}_{x\to 0}\frac{2{\sin }^2(\frac{x^2}{2})}{4\frac{x^4}{4}\frac{(4^x-1)}{x}}$
$=\frac{1}{2\log 4}=\frac{1}{4\log 2}$