fellow mathematicians. I'm currently trying to prove the following limit:
$\lim_{x \to \infty} \frac{x}{e^{x}} = 0$
Using the following figure:
I tried the following:
See the largest and the smallest triangle. The area of each is the following:
$A_{\triangle s} = \frac{((x+1)-1)(1)}{2}=\frac{x}{2}$
$A_{\triangle l} = \frac{((x+1)-x)(\frac{1}{x})}{2}=\frac{\frac{1}{x}}{2}=\frac{1}{2x}$
The asrea of under the dark curve is $lnx $. Next, I set the following inequality:
$\frac{x}{2} > lnx > \frac{1}{2x}$
Now, let $x = e$
$\frac{e}{2} > lne > \frac{1}{2e}$
Multiply by -x (inequality is changed since $-x$ is negative).
$\frac{-xe}{2} < -xlne < \frac{-x}{2e}$
Apply the logarithm multiplication property:
$\frac{-xe}{2} < lne^{-x} < \frac{-x}{2e}$
Raise $e$ to each of the terms in the inquality
$e^{\frac{-xe}{2}} < e^{lne^{-x}} < e^{\frac{-x}{2e}}$
Which in the second term simplifies to:
$e^{\frac{-xe}{2}} < e^{-x} < e^{\frac{-x}{2e}}$
Multiply everything by x
$xe^{\frac{-xe}{2}} < xe^{-x} < xe^{\frac{-x}{2e}}$
Simplify the powers:
$\frac{x}{e^{\frac{xe}{2}}} < xe^{-x} < \frac{x}{e^{\frac{x}{2e}}}$
Apply the limit:
$\lim_{x \to \infty}\frac{x}{e^{\frac{xe}{2}}} < lim_{x \to \infty}xe^{-x} < lim_{x \to \infty}\frac{x}{e^{\frac{x}{2e}}}$
Which, since the denominator grows faster, give the result:
$0 < lim_{x \to \infty}xe^{-x} < 0$
Thus
$lim_{x \to \infty}xe^{-x} = lim_{x \to \infty}\frac{x}{e^{x}} = 0$
Nonetheless, when I apply the limits in the final part I'm assuming they tend to zero, which in the first place could be assumed to "prove" the limit without having to use the figure. But I am required to use the graph, any suggestions on how to proceed in an alternative way?
For $t\ge 1$, your graph shows that $\ln t<t$.
Now let $X$ be large. Then, using $t=\sqrt X$, $$\frac{\ln X}{X}=\frac{2\ln\sqrt X}{\sqrt X\sqrt X}<\frac{2}{\sqrt X}$$
$\frac{\ln X}{X}$ therefore tends to $0$ as $X \to \infty$.
Replacing $X$ by $e^x$ then gives the result you want.