Why does the composite graph of $e^x \sin x$ appear to almost be bounded (not in the mathematical sense) by the graph $e^x$ and $-e^x$? I have seen a similar thing with the graph $e^{-x} \sin x$.
I can sketch the curve through basic analysis, but I don't see how you can deduce that it would be bounded by $e^x$?
On
look at the graph of $y = A\sin x$ where the amplitude $A$ is constant. we know that this graph bounces back and forth between $A$ and $-A$ for every $2\pi.$ with the graph of $y = e^x\sin x$ the envelope is not constant but changing and the graph bounces back and forth between the graphs of $e^x$ and $e({-x}.$
On
This is because $$|\sin x|\leq 1 \implies |f(x)\sin x|= |f(x)|\cdot|\sin x|\leq |f(x)|\cdot 1 = |f(x)|$$ In other words, we know $$-|f(x)|\leq f(x)\sin x\leq |f(x)|$$ This is true for any function $f:\mathbb R\to\mathbb R$. In your case, $e^x$ is always positive, so we can drop the absolute value bars to get $$-e^x\leq e^x\sin x \leq e^x$$
$f(x) = \sin x$ has an image of $[-1, 1]$; because it's never greater than $1$ or less than $-1$, it's never going to cause $e^x \sin x$ to be greater than $e^x$ or less than $-e^x$.