Limit of $\ln(t)t^xe^{-t}$ I want to calculate the limit: $\lim\limits_{t\to\infty}\ln(t)t^xe^{-t}$ where $x\in(0,1]$.
For $t&gt;1$ we can find the following upper bound:
$$\ln(t)t^xe^{-t}\leq t^2 e^{-t}.$$
Using the series representation of $e^{-t}$ we find another upper bound:
$$\ln(t)t^xe^{-t}\leq t^2 e^{-t}\leq t^2 \frac{3!}{t^3}=\frac{3!}{t}.$$
So it follows:
$$0\leq \lim\limits_{t\to\infty}\ln(t)t^xe^{-t}\leq \lim\limits_{t\to\infty}\frac{3!}{t}=0\Rightarrow \lim\limits_{t\to\infty}\ln(t)t^xe^{-t} =0.$$
Is this correct?
If $t\geq 1$, then $$\frac{\ln(t)}{e^t}\leq \left|\ln(t)t^xe^{-t}\right|\leq \frac{t\ln(t)}{e^{t}}.$$
Using Bernoulli L'Hospital gives the wishes result.
You can do that almost instantly with asymptotic calculus:

$\ln t=_\infty o(t^\alpha)$ for any $\alpha&gt;0$, therefore $t^x\ln t=_\infty t^xo(t)=o(t^{x+1})$,
also, $t^{x+1}=o(\mathrm e^t)$, and consequently  $\;t^x\ln  t=_\infty o(\mathrm e^t)$,
which means exactly that
$$\lim_{t\to+\infty}\frac{t^x\ln t}{\mathrm e^t}=0.$$