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: 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 $x\gt0$.
A: Let us suppose that the hypothesis is verified. Then we prove that $\lim_{x \to \infty} f(x) = 0$. The exact regularity property of $f$ is not stated by OP: here I am supposing it to be continuous and everywhere differentiable (continuity of the derivative is not required).
First we prove the following property of the function $f$:

Let $\varepsilon > 0$; then there is $x_0$ such that, for every $x \geq x_0$, if $|f(x)| < \varepsilon$ then $|f(y)| < 2\varepsilon$ for all $y \geq x$.

In order to prove this lemma let us use the hypothesis and take $x_0$ large enough so that for each $x \geq x_0$ we have
$$
\left| f(x) + \frac{f'(x)}{x} \right| < \varepsilon. \qquad (*)
$$
Now suppose that there is $y$ such that $|f(y)| \geq 2\varepsilon$ and take the smallest of such $y$, which exists by continuity and is strictly larger than $x$. Then, by continuity, it actually holds that $|f(y)| = 2\varepsilon$. We can also assume that $f(y)$ is positive; if it is not, we reason with $-f$ instead of $f$. Then $\frac{f'(y)}{y} < -\varepsilon$, so $f'(x) < 0$. This implies that just before $y$ the function $f$ is larger than $f(y) = 2\varepsilon$, but this is absurd because we took $y$ to be the smallest value greater than $x$ with that property. This establishes the lemma.
From the lemma it follows that if there are arbitrary large values of $x$ such that $f(x) = 0$, then $|f|$ is definitively smaller than $\varepsilon$ for every $\varepsilon > 0$. Then $f$ has limit zero.
So the only missing case is when $f$ is definitively positive or negative. As before the two cases are symmetrical and we cover only the positive one. Take $f$ definitively bigger than a constant $2\varepsilon > 0$. Then find $x_0$ such that for $x \geq x_0$ there holds $(*)$ above. It follows that $f'(x) < x \varepsilon \leq x_0 \varepsilon$, so that $f(x)$ must eventually descend under $2\varepsilon$, in contradiction with the assumption. This proves that in this case too $f$ must tend to zero and closes the proof.
A: Let's change the variable $t=x^2/2$ , and $\tilde f(t)=f(x)$
Then $f'(x)=\tilde f'_t x$. Thus we can rewrite 
$$
\lim_{t\to\infty}(\tilde f(t)+\tilde f^\prime(t))=0;
$$
No let's define
$$
\tilde f(t)+\tilde f'(t)=g(t)
$$
and solve $f$ as functional of $g$.
We can write 
$$
\tilde f=C e^{-t}+\int_0^t e^{\tau-t}g(t) d\tau
$$
Since $g(t)\to 0$ we have $\tilde f(t)\to 0$  
A: We know that $\lim_{x\to \infty}\left(f(x)+\frac{f'(x)}{x}\right)=0$, so for all $\varepsilon >0$ exists $y$, that for $x>y$:
$$f(x)+\frac{f'(x)}{x}<\varepsilon$$
So (because $x>0$):
$$f(x)x+f'(x)<\varepsilon x$$
Now multiply both sides by $e^{x^2/2}$:
$$f(x)xe^{x^2/2}+f'(x)e^{x^2/2}<\varepsilon x e^{x^2/2}$$
It's equal:
$$(f(x)e^{x^2/2})'<\varepsilon x e^{x^2/2}$$
And:
$$(f(x)e^{x^2/2})'-(\varepsilon e^{x^2/2})'<0$$
So function $(f(x)-\varepsilon)e^{x^2/2}$ is decreasing. So:
$$(f(x)-\varepsilon)e^{x^2/2}<(f(y)-\varepsilon)e^{y^2/2}$$
Dividing both sides by $e^{x^2/2}$:
$$f(x)-\varepsilon<(f(y)-\varepsilon)e^{y^2/2-x^2/2}$$
Let $x \to \infty$. Then:
$$\lim_{x \to \infty} f(x)-\varepsilon < \lim_{x \to \infty}e^{y^2/2-x^2/2}=0$$
The same way  $-\varepsilon<f(x)+\frac{f'(x)}{x}$ we show that $0<\lim_{x \to \infty}f(x)+\varepsilon$. It's for all $\varepsilon$, so $\lim_{x \to \infty}f(x)=0$
