If $\lim_{x \rightarrow \infty} f(x)$ is finite, is it true that $ \lim_{x \rightarrow \infty} f'(x) = 0$?

Simple counterexample: $f(x) = \frac{\sin x^2}{x}$.

UPDATE: It may seem that such an answer is an unexplicable lucky guess, but it is not. I strongly suggest looking at Brian M. Scott's answer to see why. His answer reveals exactly the reasoning that should first happen in one's head. I started thinking along the same lines, and then I just replaced those triangular bumps with $\sin x^2$ that oscillates more and more quickly as $x$ goes to infinity.

If you don’t see an example using known functions, you can think about it this way.

Start with a piecewise linear function $f$ whose graph lies along the $x$-axis except in the intervals $\left[n-\frac1{n^2},n+\frac1{n^2}\right]$ for $2\le n\in\Bbb Z$; on $\left[n-\frac1{n^2},n\right]$ it rises linearly from $0$ to $\frac1n$, and on $\left[n,n+\frac1{n^2}\right]$ it falls linearly to $0$. Clearly $\lim\limits_{n\to\infty}f(x)=0$. This function isn’t differentiable at the points $n$ and $n\pm\frac1{n^2}$ for $n\ge 2$, but it’s clearly possible to smooth out the corners. Note that the slope of the graph on $\left[n-\frac1{n^2},n\right]$ is $n$; if you smooth out the corners without changing the function values at the points $n$ and $n\pm\frac1{n^2}$, the mean value theorem will ensure that for each $n\ge 2$ there will be a point $x\in\left[n-\frac1{n^2},n\right]$ where $f\,'(x)=n$, and therefore $f\,'(x)$ will not approach $0$ as $x\to\infty$.

$$\lim_{x\to \infty} f'(x) = \lim_{x\to \infty} \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

Your claim would be true if it were allowed to switch the order of the two limits, which is not possible for all functions.