$\lim_{x\to\infty}f(x)$ exists and is finite, but $\lim_{x\to\infty}f'(x)$ does not exist 
Give an example of a function where $\lim_{x\to\infty}f(x)$ exists and is finite, but $\lim_{x\to\infty}f'(x)$ does not exist.

Does the function $f(x)=\sin(x^2)/x$ meet this criteria?
 A: For your function, $\lim_{x\to\infty}f(x)$ is clearly 0, while its derivative is
$$\frac{2x^2\cos x^2-\sin x^2}{x^2}=2\cos x^2-\frac{\sin x^2}{x^2}$$
Now consider the two sequences of real numbers $x_n=\sqrt{2n\pi}$ and $y_n=\sqrt{(2n+1)\pi}$. It is easy to show that $f'(x_n)=2$ and $f'(y_n)=-2$, so two subsequences in $f'(x)$ converge to different limits and $\lim_{x\to\infty}f'(x)$ does not exist.
A: Yes, the function does meet this criteria. For a complete elaboration :
$$\lim_{x \to \infty} \frac{\sin x^2}{x} = 0$$
since $ -1 \leq \sin x \leq 1$ and thus it will just be some number divided by infinity.
The limit of this function's derivative though, truly not exists, as :
$$f'(x) = \frac{2x^2\cos x^2-\sin x^2}{x^2}=2\cos x^2-\frac{\sin x^2}{x^2}$$
and then, the limit :
$$\lim_{x \to \infty} \bigg(2\cos x^2-\frac{\sin x^2}{x^2}\bigg) = \lim_{x \to \infty}2\cos x^2$$
Now, we know that also $-1 \leq \cos x^2 \leq 1$ but $x \to \infty$ means you cannot know the exact value of the limit, as it "fluctuates". Thus, the limit does not exist and all you can say is that :
$$\lim_{x \to \infty}2\cos x^2 \in [-2,2]$$
A more rigorous way of proving the limit does not exist :
Let $x_n = \sqrt{2n\pi}$ and $y_n = \sqrt{(2n+1)\pi}$ be two sequences of real numbers. By plugging these into the derivative expression, one can cleary see that $f'(x_n) \neq f'(y_n)$ and thus these two sequences converge to a different random number, which strictly leads to the conclusion that $\lim_{x\to\infty}f'(x)$. 
Note : Credits to Parcly Taxel for also writing the standard sequence approach as I was formulating my answer.
