1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ "\ln" gives you $\ln$ in $\TeX$. I'm not familiar with MathJax, but I've heard it is similiar for simple expressions like this. $\endgroup$ – Stefan Smith Jan 12 '14 at 17:13
  • $\begingroup$ @StefanSmith I consider it perverse notation. ;-) $\endgroup$ – egreg Jan 12 '14 at 17:14
  • $\begingroup$ @StefanSmith He meant that $\ln$ is gay and that $\log$ should be used. $\endgroup$ – Git Gud Jan 12 '14 at 17:14
  • $\begingroup$ @egreg : Got something against the French? $\ln$ is good enough for Knuth, Graham, and Patashnik! ;) $\endgroup$ – Stefan Smith Jan 14 '14 at 1:49
  • $\begingroup$ @StefanSmith Unfortunately, the dreaded ln is imposing. I'll try resisting. ;-) $\endgroup$ – egreg Jan 14 '14 at 8:26
2
$\begingroup$

Using Taylor series

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

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Git Gud Jan 12 '14 at 16:27
  • $\begingroup$ Good to see you Sami. I am just red correcting many assignments. :-( $\endgroup$ – mrs Jan 12 '14 at 16:45
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.