Need help in understanding definition of limit of a function I am using Thomas/Finney's Calculus and Analytical Geometry and it says "Given any positive radius ε about L, there exists some positive radius δ about c such that for all t within δ units of c (except t=c itself) the values of F(t) lies within ε units of L."
Here, F(t) = (f(x+t)-f(x))/(t)
Where t is the single variable and we were supposed to(which I did above) set about defining what it means for F to have a limit L as t approaches a predetermined value c.

The above thing is say if $\lim_{t \to c} F(t) = L$ if and only if for all $\epsilon > 0$, $\exists \delta > 0$ such that for all $t$, if $0<|t-c|<\delta$ then $|F(t) - L | < \epsilon$ 
Intuitively this is saying we can make the distance between $F(t)$ and $L$ arbitrarily small, by making $t$ close enough to $c$, $i.e.$, exhibiting the appropriate $\delta$.  
Here's an easy example. Take $F(t) = 3t$. Say you want to prove $$\lim_{t\to 3} F(t) = 9$$, 
This means that for an arbitrarily small real number $\epsilon > 0$, you need to find find a a positive real number $\delta$ so that if $t$ is within distance $\delta$ of $3$, the absolute difference, (distance) between $F(t)$ and $9$ is less than this small number $\epsilon$. 
Here's where you can do some scratchwork: 
The inequality: $|F(t) - 9| < \epsilon$ is:
$$
|3t-9|< \epsilon \leftrightarrow |3|\cdot |t-3| < \epsilon \leftrightarrow |t-3| < \epsilon/3
$$
So no matter what positive real number $\epsilon$ you have, if you make the distance between $t$ and $3$ less than $\delta = \epsilon/3$, i.e. if $0<|t-3|< \epsilon/3 = \delta $ then you'll have: $3|t-3| = |3t-9| = |F(t) - 9| < 3\cdot \epsilon/3 = \epsilon$.
So the limit is proved. 
