I want to calculate
$$\lim\limits_{x \to \infty} \left(\dfrac{f(x)+g(x)}{f(x)^4 + g(x)^4}\right)$$
with both functions $f$ and $g$ tending to infinity when $x$ goes to infinity. The answer is zero but is there a more mathematical solution than saying that the denominator is larger than numerator?
On
$$0\le \left(\dfrac{f(x)+g(x)}{f(x)^4 + g(x)^4}\right)=\left(\dfrac{f(x)}{f(x)^4 + g(x)^4}\right) +\left(\dfrac{g(x)}{f(x)^4 + g(x)^4}\right)$$ $$ \le\left(\dfrac{g(x)}{g(x)^4}+\dfrac{f(x)}{f(x)^4}\right) \to0 \text{ as }x\to\infty$$
Where in the last step I used $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\infty$
Thus by the sandwich therem, $$\lim\limits_{x \to \infty} \left(\dfrac{f(x)+g(x)}{f(x)^4 + g(x)^4}\right)=0$$
On
We know, that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\infty$$
Let $$h(x)=\max\{f(x),g(x)\}$$ Of course $$\lim_{x\to\infty}h(x)=\infty$$ For large enough $x$ we have: $$h(x)\leq f(x)+g(x)\leq 2h(x)$$ and $$h^4(x)\leq f^4(x)+g^4(x)\leq 2h^4(x)$$ Thus: $$\frac{f(x)+g(x)}{f^4(x)+g^4(x)}\leq \frac{2h(x)}{h^4(x)}=2\frac{1}{h^3(x)}$$ and $$\frac{f(x)+g(x)}{f^4(x)+g^4(x)}\geq \frac{h(x)}{2h^4(x)}=\frac{1}{2}\frac{1}{h^3(x)}$$
We have then $$\lim_{x\to\infty}\left(\frac{f(x)+g(x)}{f^4(x)+g^4(x)}\right)\leq 2 \lim_{x\to\infty} \frac{1}{h^3(x)}=0$$ and $$\lim_{x\to\infty}\left(\frac{f(x)+g(x)}{f^4(x)+g^4(x)}\right)\geq \frac{1}{2} \lim_{x\to\infty} \frac{1}{h^3(x)}=0$$ Thus, after the squeezing theorem $$\lim_{x\to\infty}\left(\frac{f(x)+g(x)}{f^4(x)+g^4(x)}\right)=0$$
Morally, the larger of the two functions should dominate both the numerator and the denominator. This heuristic idea is a good starting point, as it suggests to study two cases.
First, observe that by convergence to infinity both $f(x)$ and $g(x)$ are strictly positive when $x$ is large enough. Let us look at any such $x$.
If $f(x)\geq g(x)$, then $f(x)+g(x)\leq f(x)+f(x)$ and $f(x)^4 + g(x)^4\geq f(x)^4$, so $$ \frac{f(x)+g(x)}{f(x)^4 + g(x)^4} \leq \frac{2f(x)}{f(x)^4} = \frac{2}{f(x)^3}. $$ Similarly, if $g(x)\geq f(x)$, we get $$ \frac{f(x)+g(x)}{f(x)^4 + g(x)^4} \leq \frac{2}{g(x)^3}. $$ Either way, we have $$ \frac{f(x)+g(x)}{f(x)^4 + g(x)^4} \leq \frac{2}{\max\{f(x),g(x)\}^3}. $$ It may look weird to have a maximum downstairs for an upper bound like this, but it is indeed what we get.
Which of the two functions is bigger may depend on $x$. But the estimate we got holds in both cases, so it is valid for all $x$. That was the whole goal of the case-by-case estimate.
Now you just need to show that $$ \lim_{x\to\infty}\max\{f(x),g(x)\}=\infty. $$ This should be a straightforward exercise.
In fact, it is not necessary that both $f$ and $g$ go to infinity. They can even both fail to do so, as long as the maximum goes to infinity and both functions are eventually non-negative. The alternative method below doesn't work with this relaxed assumption.
This is not the only way to go, of course. I just wanted to give a method based on the idea that the bigger one dominates. You can also start with the estimate $$ \frac{f(x)+g(x)}{f(x)^4 + g(x)^4} = \frac{f(x)}{f(x)^4 + g(x)^4}+\frac{g(x)}{f(x)^4 + g(x)^4} \leq \frac{f(x)}{f(x)^4}+\frac{g(x)}{g(x)^4}. $$ There is no single correct way to do it. In fact, it's not a bad idea to see if you can do it several different ways.