Prove that $\lim_{y\to\infty}\frac{y}{t-1}e^{y(t-1)} = 0$.

Question

For the distribution $g(y)=ye^{-y}$ for $y\geq 0$ and $0$ otherwise, I have shown, through integrating by parts, that the moment generating function is:

$M_Y(t)=\int_0^\infty e^{ty}ye^{-y}dy=\left[\frac{y}{t-1}e^{y(t-1)}\right]_0 ^\infty-\int_0^\infty \frac{1}{t-1}e^{y(t-1)}=\left[\frac{y}{t-1}e^{y(t-1)}\right]_0 ^\infty +\frac{1}{(t-1)^2}$ given $t<1$.

However, I need to show that $\lim_{y\to\infty}\frac{y}{t-1}e^{y(t-1)} = 0$ in order to conclude that the $M_Y(t)=\frac{1}{(t-1)^2}$. I am having trouble evaluating the limit and can not apply the product rule for limits since the two constituent functions of it do not both have a finite limit.

Thank you for any help.

Answer 1

In THIS ANSWER, I showed using the limit definition of the exponential function and Bernoulli's Inequality that

$$e^x\ge \left(1+\frac xn\right)^n$$

for $x>-n$. Then, for $x>0$, it is easy to see that

$$e^x\ge \frac14x^2 \tag1$$

Using $(1)$, we see that for $t<1$ and $y>0$

$$\begin{align} \left|\frac{y}{t-1}e^{(t-1)y}\right|&=\frac{y}{|t-1|e^{|t-1|y}}\\\\ &\le \frac{y}{|t-1|\frac14 (t-1)^2y^2}\\\\ &=\frac4{|t-1|^3y} \end{align}$$

whence applying the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to \infty} \frac{y}{t-1}e^{(t-1)y}=0}$$

Answer 2

Change variables so $t-1=-p, p>0$. Parametrize $y=x/p$. As $x$ goes to infinity, $y$ goes to infinity, and vice-versa, so substituting this new value for $y$ into the original expression and sending $x$ to infinity preserves the original limit. It is common knowledge that $xe^{-x}$ goes to zero as $x$ goes to infinity, and multiplying it by $-p^{-2}$ (recovering the original formula, rewritten in terms of x and p) doesn't change that limit.