Existence of the $\lim _{x \rightarrow \infty}f(x)$ for a differentiable $f$

I came across this question in a graduate analysis exam:

Given a differentiable function $$f : \mathbb{R} \rightarrow \mathbb{R}$$ with $$f(1) = 2$$ and $$f'(x) = \frac{1}{x^4+f(x)^2}$$, does the limit $$\lim _{x \rightarrow \infty}f(x)$$ exist?

I don't know exactly what areas of analysis am I supposed to look into for this problem. Since we know what $$f(1)$$ is, I tried to use MVT to see what happens when $$x$$ gets very big, i.e., $$f(x) - f(1) = f(x) - 2 = (x-1)f'(t) = \frac{x-1}{t^4+f(t)^2},$$ where $$t \in (1,x)$$ is the value from the statement of the MVT. So that $$f(x) = \frac{x-1}{t^4+f(t)^2} + 2.$$ Now since $$f'$$ is positive (at least for $$x \geq 1$$), we know that $$f$$ is strictly increasing, i.e., for $$y, we have $$f(y) < f(z)$$. So we can bound $$f(x)$$ as follows: $$\frac{x-1}{x^4+f(x)^2}+2 \leq f(x) \leq \frac{x-1}{5}+2.$$ As $$x$$ gets very large, it seems like the left side will tend to $$0$$, while the right side goes to infinity, so this bound is not very useful. Are there other ways to work this problem out?

Hint: since $$f$$ is strictly increasing and always positive, $$f'(x) < 1/x^4$$ for all $$x$$. What does that imply about $$\int_1^x f'(t)dt$$?
• @KelvinLian Indeed. Now use FTC on the integral to get an upper bound for $f$. Commented Apr 6, 2022 at 16:01
• So $$f(x) - 2 \leq \int^x_1\frac{1}{t^4}dt = \frac{1}{3} - \frac{1}{3x^3},$$ unless I computed my integral wrongly. Commented Apr 6, 2022 at 16:03
• Then this implies that $\lim_{x \rightarrow \infty} f(x) \leq \frac{7}{3}$? And hence the limit exists? Commented Apr 6, 2022 at 16:05