If $f$ is differentiable in $(1,\infty )$ and $\lim_{x\to\infty }f'(x)=L<\infty $ then $\lim_{x\to\infty }f(x)=l\le\infty$?

Look at $f(x) = x^{1 \over 2} \sin(x^{1 \over 3})$. Then $f'(x) = {1 \over 2} x^{-{1 \over 2}}\sin(x^{1 \over 3}) + {1 \over 3}x^{-{1 \over 6}}\cos(x^{1 \over 3})$. The limit as $x$ goes to infinity of $f(x)$ doesn't exist; the $x^{1 \over 2}$ factor increases to infinity while the sine factor modulates it. On the other hand, the limit of $f'(x)$ is zero.