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

Question

I calculated this limit before using L'Hospital and got a result of $-1/2$, but I was wondering if we could somehow calculate this without resorting to L'Hospital and derivatives in general. Any tips and ideas will be appreciated.

Answer 1

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$$\underbrace{\lim_{x\to0}\frac12\tan x}_0-\underbrace{\lim_{x\to0}\frac{x}{\sin(2x)}}_{1/2}=-\frac12.$$

Answer 2

You can simply use the equivalence of the functions sinus and exponential at $x=0$. Thus $\sin(x)$ is equivalent to $x$ at $x=0$, and $(\sin(x))^2$ is equivalent to $x^2$ at $x=0$. In addition, $\exp(x)$ is equivalent to $1+x$ at $x=0$. Bearing this in mind, you find that $\exp( (\sin(x))^2 ) - \exp(x)$ is equivalent to $(1+x^2) - (1+x) = x(x-1)$ at $x=0$. Hence, since $\sin(2x)$ is equivalent to $2x$ at $x=0$, you find that your fraction is equivalent to $\frac{(x-1)}{2}$ at $x=0$, that is the limit is $\frac{-1}{2}$.

Answer 3

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

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

Answer 4

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