# The limit of $\ln{x} + \cot{x}$

Find the following limit:$$\lim_{x \to \ 0^+} ({\ln{x} + \cot{x}})$$ I've tried to use L'Hôpital's rule, but I can't transform the expression to the form $\frac{0}{0}$ or $\frac {\infty }{\infty}$.

Any idea?

Write the function as $$\log x+\cot x=\frac{\sin x\log x+\cos x}{\sin x}.$$ Then $$\lim_{x\to0^+}\sin x\log x=\lim_{x\to0^+}\frac{\sin x}{x}\cdot x\log x=0$$ and the previous form is not indeterminate.

(Sorry, but I can't type “ln” for the logarithm.)

• "\ln" gives you $\ln$ in $\TeX$. I'm not familiar with MathJax, but I've heard it is similiar for simple expressions like this. – Stefan Smith Jan 12 '14 at 17:13
• @StefanSmith I consider it perverse notation. ;-) – egreg Jan 12 '14 at 17:14
• @StefanSmith He meant that $\ln$ is gay and that $\log$ should be used. – Git Gud Jan 12 '14 at 17:14
• @egreg : Got something against the French? $\ln$ is good enough for Knuth, Graham, and Patashnik! ;) – Stefan Smith Jan 14 '14 at 1:49
• @StefanSmith Unfortunately, the dreaded ln is imposing. I'll try resisting. ;-) – egreg Jan 14 '14 at 8:26

Using Taylor series

$$\ln x+\cot x=\frac{\sin x\ln x+\cos x}{\sin x}\sim_0\frac{x\ln x+1}{\sin x}\sim_0\frac{1}{\sin x}\to+\infty$$

• You can only assert the first $\sim _0$ after proving that $\sin(x)\log(x)+\cos(x)\sim _0 x\log(x)+1$ and I don't think this is easy enough not to be explicitly mentioned and done. – Git Gud Jan 12 '14 at 16:27
• Good to see you Sami. I am just red correcting many assignments. :-( – mrs Jan 12 '14 at 16:45

Note that $$e^{\ln x + \cot x} = e^{\ln x}e^{\cot x} = xe^{\cot x} = \dfrac{x}{e^{-\cot x}}$$ This function is of the form $\frac{0}{0}$ as $x \to 0^+$, so you can apply L'Hôpital's rule and derive the limit of $\ln x + \cot x$ from there.