Given some information of $\limsup$ and $\liminf$ of $f$, show that $f(x)+x^{\beta}\longrightarrow\infty$ if and only if $\beta>\frac{1}{2}$. The reason of this post was another post I saw this afternoon. However, I cannot find the post anymore, so I am not able to provide the link.
The OP was about this: he knows about $$\limsup_{x\rightarrow\infty}\dfrac{f(x)}{\sqrt{2x\log\log x}}=1\ \text{and}\ \liminf_{x\rightarrow\infty}\dfrac{f(x)}{\sqrt{2x\log\log x}}=-1.$$
Then, he wants to show $f(x)+x^{\alpha}\longrightarrow\infty$ if and only if $\alpha>\frac{1}{2}$.
There was no answer but only some comments in that post. Among comments, one suggested using the fact that $$\lim_{x\rightarrow\infty}\dfrac{\log x}{x^{r}}=\infty,\ \text{for all}\ r>0,$$ so that $$\lim_{x\rightarrow\infty}(x^{r}-\sqrt{2\log\log x})=\infty\ \text{for all}\ r>0.$$
Then, using the condition of $\liminf$, writing out as $$f(x)+x^{\alpha}>x^{\alpha}+(-1-\epsilon)\sqrt{2x\log\log x},$$ it is fairly easy to prove the direction $(\Leftarrow)$.(Factor out $x^{1/2}$ and replace $r:=\alpha-\frac{1}{2}$.
However, I could not figure out $(\Rightarrow)$. Even though $r\leq 0$ implies $$\lim_{x\rightarrow\infty}(x^{r}-\sqrt{2\log\log x})=-\infty,$$ we cannot use it, since if we apply the similar argument, we would arrive at $$\liminf f(x)+x^{\alpha}>-\infty,$$ which is not useful at all. 
Then, I tried to use the condition on $\limsup$, but that only gives $\limsup f(x)+x^{\alpha}<\infty$, which is not useful either..
What should I do? thank you!
 A: Let us first consider the case $\alpha=1/2$.
Since $$\liminf_{x \to \infty} \frac{f(x)}{\sqrt{2x \log \log x}} = - 1,$$
there exists a sequence $(x_n)_{n \in \mathbb{N}}$ with $x_n \to \infty$ as $n \to \infty$ and
$$ \frac{f(x_n)}{\sqrt{2x_n\log \log x_n}} \leq -\frac{1}{2} .$$
Thus,
\begin{align*} f(x_n) + \sqrt{x_n} &= \sqrt{x_n} \left( \frac{f(x_n)}{\sqrt{x_n}}+1 \right) \\ &\leq \sqrt{x_n} \left( - \frac{1}{2} \sqrt{2 \log \log x_n} + 1 \right). \end{align*}
Since $\sqrt{x_n} \to \infty$ as $n \to \infty$ and
$$\lim_{n \to \infty} \left( - \frac{1}{2} \sqrt{2 \log \log x_n} + 1 \right) = - \infty,$$
it follows that $$\lim_{n \to \infty} f(x_n) + \sqrt{x_n} = - \infty.$$ (Note: If $(a_n)_n$, $(b_n)_n$ are real-valued sequences such that $a_n \to \infty$ and $b_n \to - \infty$, then $a_n b_n \to -\infty$.) Hence, $$\liminf_{x \to \infty} f(x) + \sqrt{x} = - \infty.$$
If $\alpha < 1/2$, then $x^{\alpha} < \sqrt{x}$ for $x \geq 1$, and hence,
$$\liminf_{x \to \infty} f(x) + x^{\alpha} \leq \liminf_{x \to \infty} f(x) + \sqrt{x} = -\infty.$$
