Why this equality holds? And how can I get from left side to the right side?

Question

I have this equality: $$\lim_{x \to 0} \frac{e^{\sin(2x)}- e^{\arcsin(x)}}{x} = \lim_{x \to 0} \frac{e^{\sin(2x)}- 1}{x} \cdot \frac{\sin(2x)}{x} - \dfrac{e^{\arcsin(x)}-1}{\arcsin(x)}.\dfrac{\arcsin(x)}{x} $$

However I can not find where the -1 come from.

If I do

$$ \lim_{x \to 0} \frac{e^{\sin(2x)}}{x} \cdot \frac{\sin(2x)}{\sin(2x)}-\frac{e^{\arcsin(x)}}{x} \cdot \frac{\arcsin(x)}{\arcsin(x)} $$

I do not get the equality above and I really do not know how can I get from the right side of the equality to the left side. Where is the -1 come from?

Answer 1

It is just \begin{align} \frac{e^{\sin 2x}-e^{\arcsin x}}{x} & = \frac{e^{\sin 2x}-1+1-e^{\arcsin x}}{x}\\ & = \frac{e^{\sin 2x}-1}{x}-\frac{e^{\arcsin x}-1}{x}, \end{align} and then $$\frac{e^{\sin 2x}-1}{x}-\frac{e^{\arcsin x}-1}{x}= \frac{e^{\sin 2x}-1}{\sin 2x}\frac{\sin 2x}{x}-\frac{e^{\arcsin x}-1}{\arcsin x}\frac{\arcsin x}{x}. $$ Now pass to the limit for $x \to 0$.

Answer 2

$e^{sin(2x)}-e^{arcsin(x)}=(e^{sin(2x)}-1)+(1-e^{arcsin(x)})=(e^{sin(2x)}-1)-(e^{arcsin(x)}-1)$

Your original expression seems to have an error. $lim_{x\to 0}\frac{e^{sin(2x)}-1}{x}\frac{sin(2x)}{x}$ should have a $2$ in the denominator.