If $f(x)+\frac{f'(x)}{x}\to0$ then $f(x)\to0$ I honestly have no idea how to solve this one. Can we use the mean value theorem?
$$\lim_{x\to \infty}\left(f(x)+\frac{f'(x)}{x}\right)=0 \implies \lim_{x\to \infty}f(x)=0$$

A general method is to consider the expression converging to zero as a function of $x$ and to express $f$ in terms of it. Here, let $$g(x)=f(x)+\frac{f'(x)}x,$$ then $$x\mathrm e^{x^2/2}g(x)=x\mathrm e^{x^2/2}f(x)+\mathrm e^{x^2/2}f'(x)=\left(\mathrm e^{x^2/2}f(x)\right)',$$ hence, for every $x\geqslant c\gt0$, 
$$\mathrm e^{x^2/2}f(x)=\mathrm e^{c^2/2}f(c)+\int_c^xt\mathrm e^{t^2/2}g(t)\,\mathrm dt.$$ Now, the usual $\varepsilon$-$\delta$ approach works, or rather, here, the $\varepsilon$-$x_0$ approach.
Fix some positive $\varepsilon$. There exists some finite $x_0\gt0$ such that, for every $x\geqslant x_0$, $|g(x)|\leqslant\varepsilon$. Thus, using $c=x_0$ in the identity above and the triangular inequality, one gets $$\mathrm e^{x^2/2}|f(x)|\leqslant\mathrm e^{x_0^2/2}|f(x_0)|+\int_{x_0}^xt\mathrm e^{t^2/2}|g(t)|\,\mathrm dt\leqslant c_0+\varepsilon\int_{x_0}^xt\mathrm e^{t^2/2}\,\mathrm dt,$$  with $c_0=\mathrm e^{x_0^2/2}|f(x_0)|$, thus, $$\mathrm e^{x^2/2}|f(x)|\leqslant c_0+\varepsilon (\mathrm e^{x^2/2}-\mathrm e^{x_0^2/2})\leqslant c_0+\varepsilon \mathrm e^{x^2/2},$$ which implies $$|f(x)|\leqslant c_0\mathrm e^{-x^2/2}+\varepsilon.$$
The first term on the RHS converges to zero hence $$\limsup_{x\to\infty}|f(x)|\leqslant\varepsilon.$$ This holds for every $\varepsilon\gt0$ hence the proof is complete.

Exercise: Follow closely the steps above to show that, for every functions $a$ and $b$ such that $a(\ )\geqslant\alpha\gt0$ uniformly and $\int^x\frac{a}b\to+\infty$ when $x\to+\infty$, if $$\lim_{x\to+\infty}a(x)f(x)+b(x)f'(x)=0,$$ then $$\lim_{x\to+\infty}f(x)=0.$$ The question above is when $a(x)=1$ and $b(x)=1/x$ for every $x\gt0$.
