Find a delta given an interval So the question is the following:
We consider $f(x) = x^2$ on the interval $[0,100]$. Let $\epsilon > 0$ be given. Find a number $\delta > 0$ so that if the length of an interval $I \in [0, 100]$ is less than delta, then $f(u) - f(l) < \epsilon$, where $f(u), f(l)$ are max and min of $f$ on $I$, respectively.
It's probably really simple but I just can't seem to wrap my head around it. Thanks!
 A: So this question is about showing the (uniform) continuity of a function on an interval. Since this seems like a homework problem, here's a somewhat simpler problem that will hopefully give some intuition:
Consider the function $f(x) = 10 x$ on $[0, 10]$.
Then, given $\varepsilon > 0$, we want to find $\delta > 0$ such that any interval of width less than $\delta$ has $f(u) - f(\ell) < \varepsilon$.
A first guess might be $\delta := \varepsilon$.
However, this doesn't work. If $\varepsilon = 1$, then $\delta = 1$, so an interval might be $[9, 10]$.
However, $f(u) - f(\ell) = f(10) - f(9) = 10 > 1 = \varepsilon$. (Why is $u = 10$, $\ell = 9$?)
Setting $\delta$ to be something less than $\frac{1}{10}\varepsilon$, will make this work out though.
In your example, you'll need to figure out where the "least well behaved" part of the function is and choose your $\delta$ based around that (hint: the steeper the function, the smaller $\delta$ has to be relative to $\varepsilon$).
A: Let me clean up your question a bit as follows.

Consider the function $f(x)=x^2$ on the closed interval $J=[0,100]$. Let $\epsilon>0$. Show that there exists $\delta >0$ such that for any closed interval $I\subset J$ with length less than $\delta$, $f(a)-f(b)<\epsilon$ where
$$
f(a)=\max\{f(r):r\in I\},\quad f(b)=\min\{f(r):r\in I\}.
$$

(This question is about the uniform continuity of $f$ on a compact interval. )
Back to your question, notice that
\begin{align}
|f(x)-f(y)|
&=|x+y||x-y|\\
&\le |x-y+2y|\cdot |x-y|\\
&\le (|x-y|+2|y|)|x-y|\\
&\le (100+2\cdot 100)|x-y|<301|x-y|,\quad x,y\in I
\end{align}
If you let $\delta=\frac{\epsilon}{301}$, then
$$
|f(x)-f(y)|< \epsilon
$$
for all $x,y\in I$ with $I$ of length $\delta$.
