I need some help with: $\lim_{x\to 0+} x^3\cdot e^{1/x}$. How to start?
I've tried substitution $(y=1/x)$ without any luck.
I would prefer not to use L'Hopitals rule and apologizes for a bad title line.
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$\begingroup$You can carry on with your substitution. In the case $y = 1/x$, then as $x \to 0^+$, $y \to \infty$, and you want to look at the limit of $$ \lim_{y\to\infty} (1/y^3) e^y = \lim_{y\to\infty} \frac{e^y}{y^3}. $$
If you know, for instance, that the exponential grows faster than any polynomial, you can avoid L'Hopital's rule.
$\endgroup$ $\begingroup$Substitute $x=\frac1t$. Then this is the same as $$\lim_{t\to+\infty}\frac{e^t}{t^3}$$ Using the wellknown inequality $e^x\ge 1+x$, we have $e^t=(e^{t/4})^4\ge (1+\frac t4)^4$, so for $t>0$ $$ \frac{e^t}{t^3}\ge\frac{(1+\frac t4)^4}{t^3}>\frac1{256}t.$$
(In short: Exponential always wins over polynomial)
$\endgroup$ $\begingroup$Hint: $e^{1/x} > \dfrac{1}{4!x^4}$.
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