hello every body
i need ahelp
We've to find the limit of:
$\displaystyle \lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)}$
It's in indeterminate form means the function is in the form of $\displaystyle \frac{0}{0}$.
So using I'Hopital's rule we get:
$\displaystyle \begin{align*}& \lim_{x \to 0}\frac{\sin{(\tan(2x))}}{\log(x+1)}\\=& \lim_{x \to 0} \frac{2\cos{(\tan(2x))}\sec^{2}(2x)}{\frac{1}{x+1} }\\=& \lim_{x \to 0} 2\cos{(\tan(2x))}\sec^{2}(2x) (x+1)\\=& (2)(1)(1)(0+1)\\=& 2......\text{[Answer]}\end{align*}$
You can check the answer here:
\lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)} - Wolfram|Alpha