Q:

Prove that for every $c>0$ there exists an $a \in [0,1)$ with $f(a) > c$

Given a continuous function $f : [0,1) \rightarrow \mathbb{R}$, $f(0) = 0$, $f(x) >0$ for $x>0$ and 
$$ \lim_{x \rightarrow 1} \frac{1}{f(x)} = 0.$$
Prove that for every $c>0$ there exists an $a \in [0,1)$ with $f(a) > c$, also prove that $f([0,1)) = [0, \infty )$.
Since $ \lim_{x \rightarrow 1} \frac{1}{f(x)} = 0$ I kow that when $x$ gets arbitrarily close to $1$, $f(x) \rightarrow \infty$. It makes sense that there exists an  $a \in [0,1)$, namely an $a$ arbitrarily close to $1$, with $f(a) > c$ for every $c > 0$. I am just not sure how to formally write a proof.
To prove $f([0,1)) = [0, \infty)$ I was thinking about taking an arbitrarily $y \in [0, \infty)$ and prove that there exists an $x \in [0,1)$ with $f(x) = y$. I am not sure how to show there exists such an $x$.
Also I didn't use the fact that $f$ is a continuous function, and I am pretty sure that I have to use it somewhere.

A:

For the first part choose $\delta $ such that $1-\delta < x <1$ implies $\frac 1 {f(x)} <\frac  1 c$. Can you finish the first part now?  
Continuous functions have Intermediate value property. This means that the range is an interval. Since $f(0)=0, f>0$ and $f$ assumes arbitrarilty large values it follows that the range has to be the interval $[0,\infty)$. 

