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) ;-(
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.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$\color{#0000ff}{\large% \lim_{x \to 0}\bracks{x^{a}\ln^{4}\pars{x}}} = \lim_{x \to \infty}\bracks{\expo{-ax}x^{4}} =\color{#0000ff}{\large% \left\lbrace% \begin{array}{lcl} +\infty & \mbox{if} & a \leq 0 \\[2mm] 0 & \mbox{if} & a > 0 \end{array}\right.} $$
