# Find the limit $\lim_\limits{x\to 0^+}{\left( e^{\frac{1}{\sin x}}-e^{\frac{1}{x}}\right)}$

Find the limit: $$\lim_\limits{x\to 0^+}{\left( e^{\frac{1}{\sin x}}-e^{\frac{1}{x}}\right)}$$

Using graph inspection, I have found the limit to be $+\infty$, but I cannot prove this in any way (I tried factorizing, using DLH)... Can anyone give a hint about that? The limit should be done without any approximations, because we haven't been taught those yet.

• With "approximations", do you include Taylor series? – Peter Woolfitt Nov 26 '15 at 18:05
• @PeterWoolfitt Yes... – Jason Nov 26 '15 at 18:06
• oh well - Taylor series seems to be the apparent approach to me – Peter Woolfitt Nov 26 '15 at 18:07
• Could you simplify $$e^a-e^b$$ ? – Stefan Nov 26 '15 at 18:07
• @Stefan I have already tried to do so in many ways by extracting $e^{1/x}$ and $e^{1/\sin x}$ in separate approaches, but both failed (even with DLH!!!)... – Jason Nov 26 '15 at 18:10

First, you can show that $$\lim_{x\to 0^+}\left(\frac{1}{\sin x}-\frac{1}{x}\right)=0.$$ This shows that $$\lim_{x\to 0^+}\frac{e^{\frac{1}{\sin x}-\frac{1}{x}}-1}{\frac{1}{\sin x}-\frac{1}{x}}=1.$$ Now, write $$e^{1/\sin x}-e^{1/x}=e^{1/x}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\frac{e^{\frac{1}{\sin x}-\frac{1}{x}}-1}{\frac{1}{\sin x}-\frac{1}{x}},$$ and letting $x\to 0^+$ gives \begin{align*}\lim_{x\to 0^+}\left(e^{1/\sin x}-e^{1/x}\right)&=\lim_{x\to 0}e^{1/x}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\lim_{x\to 0^+}\frac{e^{\frac{1}{\sin x}-\frac{1}{x}}-1}{\frac{1}{\sin x}-\frac{1}{x}}=\lim_{x\to 0^+}e^{1/x}\left(\frac{1}{\sin x}-\frac{1}{x}\right)\\ &=\lim_{x\to 0^+}\frac{e^{1/x}(x-\sin x)}{x\sin x}=\lim_{x\to 0^+}\frac{e^{1/x}(x-\sin x)}{x^2}, \end{align*} since $\lim_{x\to 0}\frac{\sin x}{x}=1$. Continuing the computation, this last limit is equal to $$\lim_{x\to 0^+}xe^{1/x}\lim_{x\to 0^+}\frac{x-\sin x}{x^3}=\frac{1}{6}\lim_{x\to 0^+}xe^{1/x}=\frac{1}{6}\lim_{u\to\infty}\frac{e^u}{u},$$ after performing the substitution $u=\frac{1}{x}$. The last limit is $\infty$, so the limit you ask is equal to $\infty$.

• It should be $x\sin x$ in the denominator, not $x^2$, initially, athough you can use $\sin x/x\to 1$ to get $x^2$ there. – Thomas Andrews Nov 26 '15 at 18:23
• Right, I'll add this. – detnvvp Nov 26 '15 at 18:23
• For the second equation you equated two limits which cannot be done because of the indeterminate form... – Jason Nov 26 '15 at 18:26
• Which one? No limit is indeterminate here. – detnvvp Nov 26 '15 at 18:27
• OK, your answer has been edited to illustrate what I had been asking! All good now. – Jason Nov 26 '15 at 18:35

By the Mean Value Theorem applied to $f(x)=e^{1/x}$ with $f'(x)=-x^{-2}e^{1/x}$, we have $$e^{1/\sin x}-e^{1/x}=f(\sin x)-f(x)=(x-\sin x)f'(\xi)=\frac{x-\sin x}{\xi^2}\cdot e^{1/\xi}$$ with $\xi$ between $x$ and $\sin x$.

We can find $a>0$ such that for all small enough positive $x$ we have $$\tag1 \sin x < x -ax^3$$ and hence $$\frac{x-\sin x}{\xi^2}>\frac{x-\sin x}{x^2}>ax.$$ Thus for small $x$ with $t:=\frac 1t$ $$e^{1/\sin x}-e^{1/x}=\frac{x-\sin x}{\xi^2}\cdot e^{1/\xi} >axe^{1/x}=a\cdot \frac{e^t}{t}\to +\infty$$ because the exponential grows stronger than any polynomial (You might use $e^x\ge 1+x$ for all $x$ $\implies e^t=(e^{t/2})^2\ge(1+\frac t2)^2=1+t+\frac14t^2$ for all $t>-2$).

How can we show $(1)$? Pick any $a$ with $0<a<\frac16$ and let $g(x)=x-ax^3-\sin x$. Then $g'(x)=1-3ax^2-\cos x$, $g''(x)=-6ax+\sin x$, $g''(x)=-6a+\cos x$, so $g'''$ is strictly decreasing from positive $1-6a$ to negative $-6a$ on $[0,\frac\pi2]$ and has a unique root $x_0\in(0,\frac\pi2)$. Thus $g''$ is strictly increasing in the interval $[0,x_0]$. Then says $g''(x)> g''(0)=0$ for all $x\in (0,x_0]$, so staht $g'$ is strictly increasing in $[0,x_0]$. Thus $g'(x)>g'(0)=0$ for all $x\in(0,x_0]$, so that $g$ is strictly increasing on $[0,x_0]$. Thus $g(x)>g(0)=0$ for $0<x\le x_0$, in other words: $(1)$.