$\newcommand{\d}{\;\mathrm{d}}$My try.
First of all, note that you have to assume that: $\forall x>0,\quad \int_0^x f(t)\d t>-\infty\; ,$ for otherwise the limit is $-\infty$ and the claim is not true.
Moreover, w.l.o.g. you can assume that $f\,$ has finite integral over each interval $[0,x]$ (because otherwise your limit is trivial).
Since $f\,$ is unbounded from above and increasing, when you choose $n\in \mathbb{N}$ you can always find $x_n\geq 0$ such that for all $x\geq x_n$, $f(x)\geq n\qquad \text{and}\qquad \int_0^x f(t)\d t\geq 0\; ;$ then for $x\geq x_n$ you get: $\frac{1}{x}\int_{x_n}^x f(t)\d t \geq \frac{n(x-x_n)}{x}\; .$ Thus: $\frac{1}{x}\int_0^x f(t)\d t \geq \frac{1}{x}\int_{x_n}^x f(t)\d t\geq \frac{n(x-x_n)}{x}$ and: $\tag{1} \liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t \geq \liminf_{x\to \infty} \frac{n(x-x_n)}{x} =n\; ;$ inequality (1) proves that $\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\ \text{d} t$ exceeds $n$ for all $n\in \mathbb{N}$, hence you necessarily have: $\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$ therefore: $\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$ as you claimed.
Just a small remark. Under your assumptions on $f$, the function $F(x):=\int_0^x f(t)\ \text{d} t$ is convex. Thus relation: $\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t=\infty$ expresses the fact that $F$ is also coercive.