Can't figure out, what to start with to define the following limit: $$ \lim_{x\rightarrow\infty} e^{-x^2} x^{x \log^2(x)} $$
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Recall that $x=e^{\log x}$. Using this you can simplify the expresion to $e^{P(x)}$ where $P(x)$ is an expression in $x$ and $\log x$. Then investigate the limit $P$ of that $P(x)$ expression. You will find $P = - \infty$, implying that your limit is $0$.
quid
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Your answer means I was going in the correct direction. However still not sure how I investigate $\lim_{x\rightarrow\infty} xlog^3x-x^2$ which I got after the modification you suggested – Dmitry Kazakov Jun 03 '14 at 16:34
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You could try to show or perhaps you can use that $x - \log^3 x \ge 1$ for sufficiently large $x$. [That it is greater 1 is is not very important; you just want that it is positive and does not tend to 0.] – quid Jun 03 '14 at 16:52