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?
 A: Use the fact that
$$3^{5 x}-2^{7 x} = e^{(5 \log{3}) x} - e^{(7 \log{2}) x}$$
and that, for small $y$
$$e^{y} = 1 + y + \cdots$$
and further, for small $z$
$$\arcsin{z} = z + \cdots$$
so that, for $x$ near $0$:
$$\frac{3^{5 x}-2^{7 x}}{\arcsin{(2 x)}-x} \sim \frac{(5 \log{3}-7 \log{2}) x}{2 x-x} = 5 \log{3}-7 \log{2}$$
A: You can write $$\frac{3^{5x}-2^{7x}}{\arcsin 2x-x}=\frac{3^{5x}-1}{\arcsin 2x-x}-\frac{2^{7x}-1}{\arcsin 2x-x}=$$$$\left(5\frac{3^{5x}-1}{5x}-7\frac{2^{7x}-1}{7x}\right)\left(2\frac{\arcsin 2x}{2x}-1\right)^{-1}.$$ Note then that all the fractions are expressed as derivatives of the functions involved, as $x\to 0$.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$\large \mbox{No H'opital, just Taylor}:$
$$
{3^{5x} - 2^{7x} \over \arcsin\pars{x} - x}
=
{%
{\bracks{1 + \ln\pars{3^{5}}x} - \bracks{1 + \ln\pars{2^{7}}x}}
\over
\pars{2x} - \pars{x}}
+
\color{#ff0000}{\Large\cdots}
=
{5\ln\pars{3} - 7\ln\pars{2}} + \color{#ff0000}{\Large\cdots}
$$
where "$\color{#ff0000}{\Large\cdots}$" terms go to zero cuando $x \to 0$.
A: 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*}
