Can one find a $g(x)$ , which is differentiable in $[a,\infty)$ and $\lim_{x\to\infty}g(x)=\infty$,but $\lim_{x\to\infty}x^2g'(x)\neq\infty$ Can one find a $g(x)$ , which is differentiable in $[a,\infty)$ and $$\lim_{x\to\infty}|g(x)|=\infty$$
but $$\lim_{x\to\infty}x^2|g'(x)|\neq\infty$$
(Or doesn't exist that limit)
 A: Let $$g(x)=x+\sin x$$ 
Then clearly, $$\lim_{x\rightarrow +\infty}|g(x)|=+\infty$$
but $$\lim_{x\rightarrow +\infty}x^2|g^\prime(x)| =\lim_{x\rightarrow +\infty}x^2|1+\cos(x)|$$ does not exist.
A: Since, if $g$ is at least absolutely continuous and $g'$ integrable,
$$|g(x)| \leqslant |g(a)| + \int_a^x |g'(t)| \, dt$$
and $|g(x)| \to \infty$, we have 
$$\int_a^\infty |g'(t)| \, dt  = \infty$$
Hence, we can rule out “nice” functions where $\lim_{x \to \infty}x^2 |g'(x)| = L < \infty$ which would imply that the integral converges with $|g'(x)| = \mathcal{O}(x^{-2})$ as $x \to \infty$.
The limit may not exist though, for example,
$$g(x) = \int_1^x \sin^2(t) \, dt$$
where $|g(x)| \to \infty$ but $x^2|g'(x)|= x^2 \sin^2 (x) $ does not converge as $x \to \infty$.
Note that $\liminf_{x \to \infty} x^2 \sin^2 (x) = 0$ and $\limsup_{x \to \infty} x^2 \sin^2 (x) = \infty$.
At this point your question is answered and we don't need to explore more pathological cases.
A: Let 
$g(x)=x\sin x+2x$,
then 
$x^2g'(x)=x^2(\sin x+2)+x^3\cos x$ .
