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.
Best Answer
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 $$\lim_{x\to0+}\bigl(x\log x\bigr)=\lim_{x\to0+}{-\log{1\over x}\over {1\over x}}=-\lim_{y\to\infty}{\log y\over y}=0\ .$$