How to show that $e^{-x/2} x^{y-1} \leq 1$ for $x$ large enough and $y\geq 1$

Question

I am wondering how to show that $e^{-x/2} x^{y-1} \leq 1$ for $x$ large enough and $y\geq 1$

I thought I could try to solve for an $x$ s.t. $e^{\frac{x}{2}} > x^{y-1}$, but if I try to take logs to get rid of the $e$, I end up with a log of $x$.

Answer 1

You cannot directly apply your idea of L'Hospital because it would not give you a constant in the end if $y$ is not an integer. So you would need to use a squeeze as well.

Another method: note that $$e^{-x/2} x^{y-1}=e^{-x/2+(y-1)\ln x}$$ So all you have to do is prove that for $a,b\geq0$, $$\lim_{x\to \infty}(a\ln x-bx)=-\infty$$ Factor by $x$ and apply L'Hopital to see that $\ln x\over x$ tends to $0$ at infinity.