Evaluate $\lim_{x \to 0+}{ (e^{1/\sin x}-e^{(1/x)})}$ I'm a high school student. For the past few hours, I was practicing the calculation of limits to improve my skills and I got stuck in a simple but complicated (for me) limit. The limit in question is

$$\lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)$$

Calculating this limit I get zero, but Wolfram says it is infinite.
Could someone help me?
$$
\begin{align}
\lim_{x\to0^+}\left(e^{\frac1{\sin(x)}}-e^{\frac1x}\right)
&amp;=\lim_{x\to0^+}e^{\frac1x}\left(e^{\frac1{\sin(x)}-\frac1x}-1\right)\\
&amp;=\lim_{x\to0^+}x\,e^{\frac1x}\lim_{x\to0}\frac{e^{\frac{x-\sin(x)}{x\sin(x)}}-1}{\frac{x-\sin(x)}{x\sin(x)}}\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{x}{\sin(x)}\\
&amp;=\infty\cdot1\cdot\frac16\cdot1\\[12pt]
&amp;=\infty
\end{align}
$$
$$\lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)= \lim_{x \to 0^+} e^{\frac{1}{x}} \left( e^\frac{x}{\sin(x)} - 1\right)$$
We know that $$\lim_{x \to 0^+} \frac{x}{\sin(x)}= \lim_{x \to 0^+} \left( \frac{\sin(x)}{x}   \right)^{-1}=1$$ and $$\lim_{x \to 0^+}\frac{1}{x}=+\infty$$
Hence:
$$ \lim_{x \to 0^+} e^{\frac{1}{x}} \left( e^\frac{x}{\sin(x)} - 1\right)=e^{+\infty}(e-1)=+\infty $$
So the initial limit is equal to:
$$ \lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)=+\infty    $$