Anything wrong with this 'proof'? Problem.
Show that $e^{-x}$ and $\sin(x)$ intersect infinitely many times.
Solution.
$\lim_{x \to o} e^{-x} = e^0 = 1$
$\lim_{x \to \infty} e^{-x} = 0$
This shows that as $e^{-x}$ goes from 0 to infinity, its range approaches 0. But this is not enough to show that they intersect infinitely many times. We must show that the range of $e^{-x}$ is contained between $1$ and $-1$ from 0 to infinity (the range of $\sin(x)$). We can do this by looking for extremas on $e^{-x}$.
$$\frac{d}{dx} e^{-x} = -e^{-x}$$
$$-e^{-x} = 0 \Rightarrow \text{False!}$$
We can therefore conclude that as $e^{-x}$ is contained within the range of $\sin(x)$ from $0$ to $\infty$, they must intersect infinitely many times.

My professor said that my proof was, and I quote: "An idea that could use more work" and only gave me 15% out of 25% for the problem on the test. He obviously wanted us to use the mean-value theorem, but as he didn't require it on the question, I thought I'd show it this way...Is there anything wrong with my proof and does it really need more work??
 A: The proof does need more work. You have shown there are no local extremes of $e^{-x}$ on $[0,\infty)$, however your conclusion, that therefore $e^{-x}$ is contained, is a total non sequitor.
The fact that a function does not have extremas has nothing at all to do with the fact that it is contained. As an example, take $f(x) = x$, which also has a nonzero derivative.
Even upon proving that, you then say that because the range of $e^{-x}$ is contained in the range of $\sin x$, this means that there are infinitely many intersections, however if you take the function $\frac{1}{2}e^{-x}$, this function has a range which is contained in the range of $e^{-x}$, but the functions never intersect.
A: Your solution is not complete as was noted by the others.
A correct solution would be:
Set $f(x)=\sin(x)-e^{-x}$. $f$ is continuous. 
Let $k\ge 0$ be an integer. Then $f(\pi/2+2\pi k)=1-e^{-(\pi/2+2\pi k)}>0$ and $f(3\pi/2+2\pi k)=-1-e^{-(3\pi/2+2\pi k)}<-1$. 
By the intermediate value theorem, $f$ has a zero in every interval $(\pi/2+2\pi k,3\pi/2+2\pi k)$ for $k\in\mathbb{N}$. That is, there are infinitely many zeros of $f$, i.e. infinitely many solutions of $\sin(x)=e^{-x}$.
A: We know the following basic facts.
1) Because of periodicity $f(x) = \sin x$ takes the values $0$ and $1$ infinitely many times at regular intervals. This means that given any number $N > 0$ there are infinitely many values of $x$ for which $x > N$ and $\sin x = 0$. And another set of infinitely many values of $x$ exist such that $x > N$ and $\sin x = 1$.
2) $\lim_{x \to \infty}e^{-x} = 0$. Note that $e^{-x} > 0$ for all $x$ and hence this limit implies that for any $\epsilon > 0$ there is a number $N > 0$ such that $0 < e^{-x} < \epsilon$ for all $x > N$. Taking $\epsilon = 1$ we see that there is an $N > 0$ such that $0 < e^{-x} < 1$ for all $x > N$.
Considering the above statements we see that for $x > N$, $e^{-x}$ lies between the bounds $0$ and $1$ (and never attains these bounds) and $f(x) = \sin x$ attains both these values $0, 1$ infinitely many times. It clearly follows by intermediate value theorem that $f(x) = \sin x$ intersects $e^{-x}$ infinitely many times. We don't need to actually figure out the points where $f(x) = \sin x$ takes the values $0, 1$. We just need to know that infinitely many such values exist. 
