I have a problem with calculating the following limits: $$\lim_{x\to0^+}\frac{\ln\sin{2x}}{\ln\cos{3x}}\\\lim_{x\to1}\frac{1-x}{\ln{x}}.$$ I wanted to use L'Hôpital's rule, but don't really know how to get $\frac{0}{0}$ or $\frac{\infty}{\infty}.$ What should I do in such cases?

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    $\begingroup$ Applying L'Hospital's rule to the second limit gives, as $x \to 1$, $$ \lim_{x \to 1}\frac{1-x}{\ln{x}}=\lim_{x \to 1}\frac{(1-x)'}{(\ln{x})'}=\lim_{x \to 1}\frac{-1}{\frac1x}=-1. $$ Concerning the first limit see @Simple Art' s answer. $\endgroup$ – Olivier Oloa Jan 2 '17 at 13:35
  • $\begingroup$ for your first limes you will get $$\infty$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 2 '17 at 13:42

For the first one that gave you infinity and zero:

By the continuity of the logarithm, we have that:

$\lim_{x \rightarrow +0}\ln{(\sin{2x})}=\ln{(\lim_{x \rightarrow +0}\sin{2x})}=\lim_{u \rightarrow +0}\ln{u}=-\infty$


$\lim_{x \rightarrow +0}\ln{(\cos{3x})=\ln{(\lim_{x \rightarrow +0}\cos{3x})}=\ln1=0}$

Because $\cos{3x}<1$ for positive $x$ close to $0$, we get that $\ln{(\cos{3x})}<0$ then. So:

$\lim_{x \rightarrow +0}\frac{\ln{(\sin{2x})}}{\ln{(\cos{3x})}}=(-\infty)(-\infty)=+\infty.$


The following are not indeterminate forms:



L'Hospital's rule is not applicable, and I recommend you check why the above should be true.

  • $\begingroup$ Yeah, I thought so about that $\frac{0}{\infty}$, but the problem is that the answer for that limit is $-1$, not $0$. $\endgroup$ – R.K. Jan 2 '17 at 13:36
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    $\begingroup$ @R.K. Well, the second limit is clearly not $\frac0\infty$... $\endgroup$ – Simply Beautiful Art Jan 2 '17 at 13:36
  • $\begingroup$ $\ln 1$ is $0$, not $\infty$. $\endgroup$ – MPW Jan 2 '17 at 13:37
  • $\begingroup$ Yeah, it's not. My mistake. $\endgroup$ – R.K. Jan 2 '17 at 13:37

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