$\frac1x\le f'(x)\le x$ implies $\lim_{x\to\infty}\ f(x)=+\infty$ $f:~[1,+\infty)\to \mathbb{R} $ is a differentiable function such that $$\frac1x\le f'(x)\le x$$
Show that $\lim_{x\to\infty}\ f(x)=+\infty$
My idea: $f$ is increasing, it remains to show it is unbounded.
Integrals shouldn't be used,problem is from the section on differential calculus.
 A: Integration here is a much more natural way to prove but I can propose a method based on Lagrange theorem.
For $1 < a < b$ consider equality 
$$
\frac{f(b) - f(a)}{b - a} = f'(x)
$$ for some $x \in [a,b]$. From here you obtain inequality
$$
f(b) \ge f(a) + \frac{b - a}{b}
$$ since $f'(x) \ge \frac{1}{b}$. Now let's try $b = 2a$. You can get
$$f(2a) \ge f(a) + \frac{1}{2}$$
for all $a > 1$. So, your function, obviously, is unbounded.
A: Expanding upon comment of $@sera$, by mean value theorem, we get,
$$f(n+1)-f(n)=f'(c)\ge\frac{1}{c}\ge\frac{1}{n+1}$$
$$\implies f(n+1)-f(n)\ge \frac{1}{n+1}$$
Sum both sides from $n=1$ to $n=m$,
$$\sum_{n=1}^m\Big(f(n+1)-f(n)\Big)\ge \sum_{n=1}^m\frac{1}{n+1}$$
$$\implies f(m+1)\ge f(1)+\sum_{n=2}^{m+1}\frac{1}{n}$$
The unboundedness of $f$ now follows from divergence of harmonic sum.
A: We have that $\frac{1}{x} \leq f'(x)$, i.e., $0 \leq f'(x) - \frac{1}{x}$. This means that the function $f(x) - \ln{x}$ is monotonically increasing, i.e. for $x \leq y$ it follows that $f(x) - \ln{x} \leq f(y) - \ln{y}$, or $f(x)+\ln{y}-\ln{x} \leq f(y)$. For an arbitrary $x > 0$, let $y := 2x$, then it follows that $f(x) + \ln{2} \leq f(2x)$ for all $x > 0$. Lastly, let $x_n := 2^n$. It now follows by induction that $f(x_n) \geq f(1) + n\ln{2} \rightarrow \infty$ and then, the claim follows through the monotony of $f$.
A: Here is a more general proposition:

If $f,g:[1,\infty)\to\Bbb R$ are two differentiable functions such that $g'(x)\le f'(x)$ for all $x$, then $g(x)-g(1)\le f(x)-f(1)$ for all $x$.

Proof: We first prove the case $g'(x)=0$. You may already know that function with zero derivative is constant (proved using mean value theorem). So $g(x)-g(1)$ is constant $0$. Mean value theorem says that $$\frac{f(x)-f(1)}{x-1}=f'(c)\ge g'(c)=0,$$ meaning $f(x)-f(1)\ge0=g(x)-g(1)$.
So the following statement is true:

If $F:[1,\infty)\to\Bbb R$ is differentiable such that $0\le F'(x)$ for all $x$, then $0\le F(x)-F(1)$ for all $x$.

Now if you are given two differentiable functions $f,g$ with $g'(x)\le f'(x)$, put $F(x)=f(x)-g(x)$, verify that $0\le F'(x)$ holds, and obtain the result $0\le F(x)-F(1)$ which is the same as $g(x)-g(1)\le f(x)-f(1)$. Q.E.D.
Going back to your question, $\frac{1}{x}$ is the derivative of $\log(x)$ which you can verify. The logarithmic function grows to positive infinity as $x\to\infty$ and so must $f(x)$ by comparison.
