# Limit of $(\cos x-1)e^{\cot x}$ as $x$ approaches $0^{+}$

Problem: Compute the limit $$\lim_{x\to 0^{+}}(\cos x-1)e^{\cot x}$$ My solution: $$\lim_{x\to 0^{+}}(\cos x-1)e^{\cot x}=\lim_{x\to 0^{+}}\frac{\cos x-1}{x^2}x^2e^{\cot x}=-\frac 12\lim_{x\to 0^{+}}x^2e^{\cot x}$$ and then $$\lim_{x\to 0^{+}}x^2e^{\cot x}=\lim_{x\to 0^{+}}\left(x^2+x^2\cot x+\frac 12x^2\cot^2 x+\sum_{n=3}^{\infty}\frac{x^2\cot^n x}{n!}\right)$$ The first three terms have limit $$0$$ and $$\lim_{x\to 0^{+}}x^2\cot^n x=\infty$$ for $$n\geq 3$$, so the last limit is $$+\infty$$ and the limit we want is $$-\infty$$.

I'm looking for different solutions, perhaps using squeeze theorem or a different way to expand with Taylor series (L'Hopital's rule doesn't seem to work nicely here).

• Hint: $$e^{\cot{(x)}}\gt\frac1{x^3} \text{ for }0\lt x\lt 0.1$$ – Peter Foreman Nov 24 '19 at 18:12
• @PeterForeman How do we get that inequality? – bjorn93 Nov 24 '19 at 18:16
• Use the facts that $\cot{(x)}\gt\frac1x-1\gt0$ for small enough $x$ and $e^x\gt (x+1)^3$ for large enough $x$ for example. – Peter Foreman Nov 24 '19 at 18:29
• The third term actually has limit $\frac12\ne0$, but for your argument it doesn't matter. – J.G. Nov 24 '19 at 18:59

We have

$$(\cos x-1)e^{\cot x}= \frac{\cos x-1}{x^2}x^2e^{\cot x}\to -\infty$$

indeed by standard limits

• $$\frac{\cos x-1}{x^2}\to -\frac12$$

• $$x^2e^{\cot x}= \frac{x^2}{\tan^2x}\cdot \frac{e^{\cot x}}{\cot^2x}\to 1\cdot \infty$$

A possible way is to substitute $$x= \arcsin t$$ and consider $$t \to 0^+$$.

Just for convenience flipping the sign in the first expression we get

$$\begin{eqnarray*} (1-\cos x)e^{\cot x} & \stackrel{x= \arcsin t}{=} & \left(1-\sqrt{1-t^2}\right)e^{\frac{\sqrt{1-t^2}}{t}} \\ & \stackrel{u>0: e^u> \frac{u^4}{24}\; (series\; expansion)}{>} & \frac{t^2}{1+\sqrt{1-t^2}}\cdot\frac{(1-t^2)^2}{24t^4}\\ & = & \frac{(1-t^2)^2}{1+\sqrt{1-t^2}}\frac{1}{24t^2} \\ & \stackrel{t\to 0^+}{\longrightarrow} & +\infty \end{eqnarray*}$$

Hence, $$\boxed{\lim_{x\to 0^{+}}(\cos x-1)e^{\cot x} = -\infty}$$.

We can use equivalents:

• As $$\lim_{x\to 0} \smash{\Bigl(\cot x-\dfrac 1x\Bigr)}=0$$, we have $$\mathrm e^{\cot x}\sim_{0^+}\mathrm e^{\tfrac1x}$$
• Also $$\cos x-1\sim_0-\dfrac{x^2}2$$, so $$(\cos x-1)\mathrm e^{\cot x}\sim_{0^+} -\frac{x^2}2\mathrm e^{\tfrac1x} =-\frac{\mathrm e^{\tfrac1x}}{\cfrac 2{x^2}}\to -\infty.$$