How do I evaluate this limit with an integral function? Could anybody give me some pointers on how do I evaluate this limit:
$$\lim_{t \to +\infty} \frac{\int_0^t x^9e^{-x^2}\,dx}{\int_0^t x^7e^{-x^2}\,dx}$$
I'm not asking for the complete solution, just a hint to point me in the right direction. I appreciate your help. Thanks.
 A: We don't need to evaluate the integrals. 
Integrating by parts we obtain
\begin{align}
\int_0^{\infty} x^9 e^{-x^2}dx&=-\frac{1}{2}\int_0^{\infty} x^8 (-2x)e^{-x^2}dx\\
&=-\frac{1}{2}\left[x^8 e^{-x^2} \right]_0^{\infty}+\frac{8}{2}\int_0^{\infty} x^7 e^{-x^2}dx\\
&=4\int_0^{\infty} x^7 e^{-x^2}dx.
\end{align}
So
$$
\frac{\int_0^{\infty} x^9 e^{-x^2}dx}{\int_0^{\infty} x^7 e^{-x^2}dx}=4.
$$
I think the hint "integration by parts" would not be enough, but I left some details for you to make complete the calculations.
A: $\newcommand{\+}{^{\dagger}}
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With $\ds{\mu > 0}$:
\begin{align}
&\color{#00f}{{\ds{\int_{0}^{\infty}x^{9}\expo{-x^{2}}\,\dd x} \over  \ds{\int_{0}^{\infty}x^{7}\expo{-x^{2}}\,\dd x}}}
=\left.-\,\partiald{}{\mu}\ln\pars{\int_{0}^{\infty}x^{7}\expo{-\mu x^{2}}\,\dd x}
\right\vert_{\mu\ =\ 1}
\\[3mm]&=\left.-\,\partiald{}{\mu}\ln\pars{\mu^{-4}\int_{0}^{\infty}x^{7}
\expo{-x^{2}}\,\dd x}\right\vert_{\mu\ =\ 1}
=\left.-\,\partiald{\ln\pars{\mu^{-4}}}{\mu}\right\vert_{\mu\ =\ 1}
=\left.{4 \over \mu}\right\vert_{\mu\ =\ 1}
=\color{#00f}{\Large 4}
\end{align}
A: $\frac{\int_0^tx^9e^{-x^2}dx}{\int_0^tx^7e^{-x^2}}=\frac{-\frac{1}{2}(e^{-t^2}(t^8+4t^6+12t^4+24t^2+24)+24)}{-\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6)+6)}=\frac{\frac{1}{2}e^{-t^2}(t^8+4t^6+12t^4+24t^2+24+24e^{t^2})}{\frac{1}{2}(e^{-t^2}(t^6+3t^4+6t^2+6+6e^{t^2}))}=\frac{t^8+4t^6+12t^4+24t^2+24+24e^{t^2}}{t^6+3t^4+6t^2+6+6e^{t^2}}=\frac{e^{-t^2}(t^8+4t^6+12t^4+24t^2+24)+24}{e^{-t^2}(t^6+3t^4+6t^2+6)+6}\to 4\,\,\,,\,as\,t\to\infty$ 
