Proving that inequality

Question

I was trying to show :

$t>0 \ \ $ there exists a $\ x_0\gt0 \ $ for every $x\geq x_0$ $\ \ \ x^ne^{-tx}\leq e^{-tx/2} \ $ ($n \in \mathbb N$)

It is clear that $e^{-tx}\leq e^{-tx/2} $ since $e^x$ is an increasing function but I could not determine $x_0$ in no way.

I appreciate for any help.

Thanks in advance

Answer 1

Note that$$\lim_{x\to\infty}\frac{x^ne^{-tx}}{e^{-tx/2}}=\lim_{x\to\infty}\frac{x^n}{e^{tx/2}}=0$$and therefore $\dfrac{x^ne^{-tx}}{e^{-tx/2}}<1$ if $x$ is large enough.

Answer 2

Taking the logarithm on both sides we get $$n\ln(x)-tx\le -\frac{tx}{2}$$ so we have $$n\ln(x)\le tx$$ for $x>0$ we get $$\frac{\ln(x)}{x}\le \frac{t}{n}$$

Answer 3

Your question is equivalent to inspect if $$e^{(tx)/2}\geq x^n$$. But it is well known that exponentials grows faster then polynomials, $$\lim_{x\to\infty} \frac{x^n}{e^{(tx)/2}}=0.$$

It clearly implies that for all $x$ sufficiently large enough, the term $\frac{x^n}{e^{(tx)/2}}$ is small enough so that $1\geq \frac{x^n}{e^{(tx)/2}}$ holds.