Finding the limit of $f$ given $f(0)>0, f(x) \leq f'(x).$ Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function, with $f(0) > 0$, so $f(x) \leq f'(x)$. 
Prove that $$\lim_{x\rightarrow \infty } f(x)=\infty.$$
I tried to use lagrange, but this doesn't lead me to the conclusion. 
 A: $y = e^x$ is a solution to the ODE $y(x) = y'(x)$.
We know that $f(x) \leq f'(x)$, so hopefully we should be able to prove that $f(x)$ grows faster than $e^x$. This will be sufficient to show that $\lim_{x \to \infty} e^x = \infty$.
So let us define a new function $g : \mathbb R \to \mathbb R$ by
$$ g(x) = \frac{f(x)}{e^{x}}.$$
Note that $g$ is differentiable (Why?), and 
$$ g'(x) = \frac{f'(x)}{e^{x}} - \frac{f(x)}{e^{x}} \geq 0.$$
Therefore, $g$ is an increasing function.
Can you finish off from here?
A: From $f(0)>0$ and $f'(0)\geq f(0)>0$, we get $f'(0)\geq0$. If $f$ is ever going to decrease, we need a maximum $c$ with $f'(c)=0$; but then we would have $f(c)\leq0$, which is impossible. It follows that $f(x)\geq0$ for all $x$. 
So $\lim_{x\to\infty}f(x)$ exists, either as a number or $\infty$. If $\lim_{x\to\infty}f(x)=L<\infty$, this implies that, eventually, $f'(x)>L/2$. So, for $x$ big enough, 
$$
f(x)-f(x-1)=\int_{x-1}^xf'(t)\,dt>\frac L2,
$$
contradicting the existence of the limit of $f$. So the only possibility is that $\lim_{x\to\infty}f(x)=\infty$. 
