$f(x)=e^{-x}\sin x$ bounded for nonnegative $x$

Question

How do I show that the function $f(x)=e^{-x}\sin x$ and its derivative $f'(x)=e^{-x}(\cos x-\sin x)$ are bounded for on the set $\{x \ge 0\}$?

Probably I need to use that $|f(x)|\rightarrow 0$ and $|f'(x)|\rightarrow 0$ as $x\rightarrow \infty$. For example, since $|f|\rightarrow 0$, there is $M>0$ such that $|x| > M \implies |f(x)|\lt\epsilon$. Consider $[0,M]$. $f$ is continuous so attains its maximum there and hence bounded. The same argument seems to work for $f'$.

Is this argument correct?

Answer 1

Yes your argument is correct! But you can also prove the statement in an elementary manner:

Use that $g_1(x)=e^{-x}$ is strictly decreasing therefore for $x\geq 0$:

$g_1(x)\leq g_1(0)=1$.

Of course the $g_2(x)=\sin x$ is also bounded : $|\sin x|\leq1$ for all $x$.

Since $f=g_1\cdot g_2$ we have that $|f|\leq g_1(0)=1$.

Similarly ,for $g_3(x)=\cos x -\sin x $ we have $|g_3(x)|\leq 2$ and we work as before.

Answer 2

HINT

Note that we are dealing with product of bounded functions, indeed

$$0\le e^{-x}\le1 \quad x\ge0$$

$$0\le \sin x \le1$$

$$-\sqrt 2\le \cos x - \sin x=\sqrt 2 \sin\left(x+\frac{\pi}{4}\right) \le \sqrt 2 $$