I'm curious about whether these two limits are the same (well I know they are equal since Wolfram Alpha confirms it, but I want to know whether the reasoning is justified):

$$ \lim_{x\rightarrow \infty} \frac{\ln{x}}{x} \;\; \text{ is equivalent to } \;\; \lim_{x\rightarrow 0}\;x\ln{x} $$

I've already found that: $$ \lim_{x\rightarrow \infty}\frac{\ln{x}}{x} =0 $$ and then tried to use this to find the second limit: $$ \lim_{x\rightarrow \infty}\frac{\ln{x}}{x} = \lim_{x\rightarrow \infty}\; \frac{1}{x}\times \ln{ \left( \frac{1}{x} \right)^{-1} }= -\lim_{x\rightarrow \infty}\; \frac{1}{x}\times \ln{ \left( \frac{1}{x} \right) } $$ since $1/x$ tends to $0$ as $x$ tends to $\infty$ then I let: $y=1/x$ and thus: $$= -\lim_{1/y\rightarrow\infty} \; y\ln{y} $$ and given that $1/y \rightarrow \infty$ would imply that $y\rightarrow 0$: $$=-\lim_{y\rightarrow 0}y\ln{y}$$ So the limit would be: $$-\lim_{y\rightarrow 0}y\ln{y}=\lim_{x\rightarrow \infty}\frac{\ln{x}}{x} =0$$

I'm just not quite sure whether this is strictly correct, since it seems to me that $1/x$ would approach zero at a different 'rate' than $x$ would and thus the limits wouldn't necessarily have to be the same.

  • 2
    $\begingroup$ It's perfectly correct. The definition of a limit doesn't mention the notion of ratefor approaching $0$, and anyway the term approach is purely metaphorical: in a limit, nothing really moves to or approach whatsoever. $\endgroup$
    – Bernard
    Apr 24, 2015 at 12:08
  • $\begingroup$ Note however that the two limits "approach" 0 from different directions. In practical applications, you may never reach infinity so you are likely to "approach" your limit... $\endgroup$
    – citronrose
    Apr 24, 2015 at 12:09
  • $\begingroup$ @Bernard Thanks! That makes sense, I just was just trying to think of reasons it could be wrong, and the idea of the 'rates' was the best I could come up with. Again Thanks for you help! $\endgroup$
    – Jay
    Apr 24, 2015 at 12:13
  • $\begingroup$ @Bernard Would you mind taking a look at my question? $\endgroup$
    – Jean-Paul
    Sep 16, 2015 at 7:40

1 Answer 1


There is a general theorem about nested limits $\lim_{x\to\xi}g\bigl(f(x)\bigr)$:

If the limits $$\lim_{x\to\xi} f(x)=\eta, \qquad \lim_{y\to\eta} g(y)$$ exist, and $g$ is continuous at $\eta$, if $f$ assumes this value at all, then $$\lim_{x\to\xi}g\bigl(f(x)\bigr)=\lim_{y\to\eta} g(y)\ .$$ Here $\xi$, $\eta$, and the limits in question are allowed to be infinite.

In the case at hand it follows that $$\lim_{x\to0+}\bigl(x\log x\bigr)=\lim_{x\to0+}{-\log{1\over x}\over {1\over x}}=-\lim_{y\to\infty}{\log y\over y}=0\ .$$


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