Limit of a sequence with ln. even though it's not actually a homework rather than some training for myself, I'm posting it with tag "Homework":
$ \ln(x)^4 $ means: $ \left( \ln (x) \right)^4 $
What is $\lim_{x\to 0} (x^a \cdot \ln(x)^4)$ ? I am not allowed to use L-Hospital.
I tried writing it like:
$\lim_{x\to 0} (x^a \cdot ln(x)^4) = \lim_{x \to 0} (e^{a\cdot \ln(x)} * e^{\ln(\ln(x)^4)}) = \lim_{x \to 0} (e^{a\cdot \ln(x) + \ln(\ln(x)^4})$ ... Well, I'm really not sure whether I made the right choice(s) ;-(
 A: In the case when $a\le0$, the computation is easy, so we can assume $a>0$.
One can simplify the computation by considering first
$$
\lim_{x\to0}x^b\ln x
$$
where $b=a/4$. If we set $\ln x=-t$, we have $x^b=e^{-bt}$ and the limit becomes
$$
-\lim_{t\to\infty}te^{-bt}=
-\lim_{t\to\infty}\frac{t}{e^{bt}}=-\frac{1}{b}\lim_{t\to\infty}\frac{bt}{e^{bt}}
=-\frac{1}{b}\lim_{u\to\infty}\frac{u}{e^u}.
$$
This is a known limit.
A: $\newcommand{\+}{^{\dagger}}%
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$$\color{#0000ff}{\large%
\lim_{x \to 0}\bracks{x^{a}\ln^{4}\pars{x}}}
=
\lim_{x \to \infty}\bracks{\expo{-ax}x^{4}}
=\color{#0000ff}{\large%
\left\lbrace%
\begin{array}{lcl}
+\infty & \mbox{if} & a \leq 0
\\[2mm]
0 & \mbox{if} & a > 0
\end{array}\right.}
$$
