Can you help me with the limit of a function up and down bounded? 
If ${2x-1\le f(x)\le x^2}$ for ${0<x<3}$.  Find $\lim_{x\to 1}f(x)$.
  How can I solve this?

 A: Since $f(x)\le x^2$, then:
$$\lim_{x \to 1} f(x) \leq \lim_{x \to 1} x^2 = 1.$$
On the other hand, since $f(x) \geq 2x-1$, then:
$$\lim_{x \to 1} f(x) \geq \lim_{x \to 1} 2x-1 = 1.$$
As a consequence, we get that:
$$1 \leq \lim_{x \to 1} f(x) \leq 1.$$
Then:
$$\lim_{x \to 1} f(x) = 1.$$

This is a good example for the squeeze theorem. Indeed, if $a(x) \leq f(x) \leq b(x)$, then:
$$\lim_{x \to k} a(x) \leq \lim_{x \to k} f(x) \leq \lim_{x \to k} b(x).$$
Then, if $\lim_{x \to k} a(x) = \lim_{x \to k} b(x) = q$, we can conclude that:
$$\lim_{x \to k} f(x) = q.$$
A: Look at the problem in this way: you know nothing about $f(x)$ (since you're not provided with any definition of $f$) except that for every $x$ in the interval $(0,3)$, $f(x)$ is lower than $x^2$ and at the same time greater than $2x-1$. Then, how would you find the limit of $f$ as $x$ approaches $1$ (or any other point within the mentioned interval)?
It's simple. You just need to check the boundary functions ($x^2$ and $2x-1$) to see whether they're both having the same limit at that point or not. If so, then $f$ has to stuck with that, too, because $f$ lies between them. This is what the Squeeze theorem states.
You can see $$\lim_{x\to1}2x-1=1$$ and $$\lim_{x\to1}x^2=1$$ Therefore, it has to be the same for $f$, i.e., $$\lim_{x\to1}f(x)=1$$
