# Comparing $\delta$ to $x-a$ in limit as $x$ approaches $a$

I have an Epsilon Delta definition of limits in my textbook that says:

The limit of $f(x)$ as $x$ goes to $a$ is $L$, if for every $\epsilon$ there is a $\delta$ so that whenever

$0<|x-a|<\delta$ we have $|f(x)-L|< \epsilon$

I have a hard time understanding the meaing of $\delta$ in Epsilon Delta definition of limits.

As $x$ approaches $a$, we have a difference $|x-a|$.

Isn't that value, $|x-a|$, the meaning of $\delta$?

• What do you mean by "value of $\delta$"? – GoodDeeds Feb 27 '16 at 19:04
• @GoodDeeds I mean isn't $\delta$ defined by $x-a$? Or $\delta = x - a$ – Eric Johnson Feb 27 '16 at 19:05
• No. It means that there is some $\delta>0$ such that for all $x$ that satisfy $0 < |x-a| < \delta$ that the condition holds. For example, take $f(x) = x^2$ and $a=0$ (and so $L=0$). Then, given $\epsilon>0$, choose $\delta= \sqrt{\epsilon}$ and then if $0<|x-a| < \delta$, we have $|f(x)-L| < \epsilon$. However, we could have chosen $\delta = {1 \over 2} \sqrt{\epsilon}$ or some other formula as long as the condition holds. – copper.hat Feb 27 '16 at 19:06
• $|x-a|<\delta$ means that $x$ is in the interval $(a-\delta,a+\delta)$. So, the statement means that there exists such an interval around $a$ such that for each $x$ in this interval we have that $f(x)$ is $\varepsilon$-close to $L$. – Friedrich Philipp Feb 27 '16 at 19:09
• It might help to draw a little picture. Then the $\epsilon$ defines a horizontal slab about $L$ on the $y$-axis. You are trying to find a vertical slab (defined by $\delta$) around $x$ so that the graph of $f$ doesn't touch the bottom or top of the rectangle defined by the intersection. – copper.hat Feb 27 '16 at 19:09

You have to think of $\epsilon$ and $\delta$ as radius of certain neigborhoods of $L$ and $a$ respectivly. The idea is that no matter how close you "ask" the function to be to L (how small you chose $\epsilon$ to be), you can always find some $\delta$ that verifies that whenever the distance from $x$ to $a$ is less than $\delta$ you will get that $f(x)$ is as close as you wanted to $L$ (the distance from $f(x)$ to $L$ is less than $\epsilon$)