Calculating the limit $ \lim_{x \to 0} \frac{e^{\sin^2(x)} - e^x}{\sin(2x)}$ without L'Hospital's rule

You want$$\underbrace{\lim_{x\to0}e^x}_{1}\cdot\underbrace{\lim_{x\to0}\frac{e^{\sin^2x-x}-1}{\sin^2x-x}}_{1}\cdot\lim_{x\to0}\frac{\sin^2x-x}{\sin(2x)}$$(the second limit uses $\lim_{y\to0}\frac{e^y-1}{y}=1$). The last limit is$$\underbrace{\lim_{x\to0}\frac12\tan x}_0-\underbrace{\lim_{x\to0}\frac{x}{\sin(2x)}}_{1/2}=-\frac12.$$

Just to give an alternative to J.G.'s approach, we have

$$\begin{align} {e^{\sin^2x}-e^x\over\sin2x} &={(e^{\sin^2x}-1)-(e^x-1)\over2\sin x\cos x}\\ &={1\over2\cos x}\left({e^{\sin^2x}-1\over\sin^2x}\sin x-{e^x-1\over x}\cdot{x\over\sin x} \right)\\ &\to{1\over2\cdot1}\left(1\cdot0-1\cdot1 \right)\\ &=-{1\over2} \end{align}$$

Anyway, no need for L'Hospital's rule here.

Just use Maclaurin's expansions at the relevant orders and compose them where required:

• $\mathrm e^x=1+x+o(x)$;
• $\sin^2x=x^2+o(x^2)$, so $\mathrm e^{\sin^2x}=1+x^2+o(x^2)=1+o(x)$;
• $\sin 2x=2x+o(x)$.

Therefore $$\frac{\mathrm e^{\sin^2x}-\mathrm e^x}{ {\sin2x}}=\frac{-x+o(x)}{2x+o(x)}=-\frac12+o(1).$$