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?
 A: $$
\begin{align}
\lim_{x\to0^+}\left(e^{\frac1{\sin(x)}}-e^{\frac1x}\right)
&=\lim_{x\to0^+}e^{\frac1x}\left(e^{\frac1{\sin(x)}-\frac1x}-1\right)\\
&=\lim_{x\to0^+}x\,e^{\frac1x}\lim_{x\to0}\frac{e^{\frac{x-\sin(x)}{x\sin(x)}}-1}{\frac{x-\sin(x)}{x\sin(x)}}\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{x}{\sin(x)}\\
&=\infty\cdot1\cdot\frac16\cdot1\\[12pt]
&=\infty
\end{align}
$$
A: Take $e^{1/x}$ as common factor and note that the remaining factor can be written as $$\dfrac{\exp\left(\dfrac{1}{\sin x} - \dfrac{1}{x} \right) - 1} {\dfrac{1}{\sin x} - \dfrac{1}{x}}\cdot\left(\frac{1}{\sin x} - \frac{1}{x}\right)$$ The first factor above tends to $1$ (why? because the second factor tends to $0$) and hence the desired limit is equal to the limit of $$e^{1/x}x\cdot\frac{x-\sin x} {x^3}\cdot\frac{x}{\sin x} $$ The last two factors give us limit $1/6$ and hence the desired limit is equal to the limit of $$\frac{1}{6}\frac{e^{1/x}}{1/x}$$ and clearly the exponential term tends to $\infty$ much faster than $1/x$ resulting in the limit being $\infty$.
A: Invoking the inequality $\sin x\lt x$ for $x\gt0$, we have
$$\begin{align}
e^{1/\sin x}-e^{1/x}
&=\int_{1/x}^{1/\sin x}e^udu\\
&={1\over x}\int_1^{x/\sin x}e^{t/x}dt\\
&\gt{1\over x}\left({x\over\sin x}-1\right)e^{1/x}\\
&={x-\sin x\over x^3}\cdot{x\over\sin x}\cdot{e^{1/x}\over1/x}
\end{align}$$
The three pieces give either familiar or straightforward L'Hopital limits:
$$\lim_{x\to0}{x-\sin x\over x^3}={1\over6}$$
$$\lim_{x\to0}{x\over\sin x}=1$$
and
$$\lim_{x\to0^+}{e^{1/x}\over1/x}=\lim_{u\to\infty}{e^u\over u}=\infty$$
Since $e^{1/\sin x}-e^{1/x}$ is greater than the product of the three pieces, we have
$$\lim_{x\to0^+}\left(e^{1/\sin x}-e^{1/x}\right)=\infty$$
as well.
A: We have that


*

*$\sin x=x-\frac16x^3+o(x^3)$

*$\frac{1}{\sin x}=(x-\frac16x^3+o(x^3))^{-1}=\frac1x (1-\frac16x^2+o(x^2))^{-1}=\frac1x+\frac16x+o(x)$


and thus
$${e^\frac{1}{\sin x}-e^\frac{1}{x}}=e^\frac{1}{x}\left(e^{\frac16x+o(x)}-1\right)=e^\frac{1}{x}\left(\frac16x+o(x)\right)=xe^\frac{1}{x}\left(\frac16+o(1)\right)\to \infty$$
indeed by $y=\frac1x\to \infty$
$$xe^\frac{1}{x}=\frac{e^y}{y}\to \infty$$
A: $$\lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)= \lim_{x \to 0^+} e^{\frac{1}{x}} \left( e^\frac{x}{\sin(x)} - 1\right)$$
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:
$$ \lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)=+\infty    $$
