Can't solve this problem $lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x}$(Without using L'Hospital's rule) I don't see the way to  solve this limit. 
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} $$
My attempt is 
1) Divide the numerator by $3^{5x}$
$$\lim_{x \to 0}\frac{3^{5x}-2^{7x}}{\arcsin\left(2x\right) - x} = \lim_{x \to 0}  \frac {1- \left(2/3 \right)^{5x}\,\,2^{2x}}{\arcsin\left(2x\right)-x} $$
And I am unable to do next step. So how this limit can be solved?

Using L'Hospital's rule, we find 
\begin{align*}
\lim_{x \to 0} \frac{3^{5x} - 2^{7x}}{\arcsin{2x} - x} &= \lim_{x \to 0} \frac{3^{5x} (5 \ln{3}) - 2^{7x}(7 \ln{2})}{\frac{2}{\sqrt{1 - (2x)^2}} - 1} \\
&= \frac{3^0 (5 \ln{3}) - 2^0 (7 \ln{2})}{2 - 1} \\
&= 5 \ln{3} - 7 \ln{2}
\end{align*}
