# For the epsilon-delta definition of limits, can you always pick $\delta$ to be $\epsilon/4$?

I just want to get a better grasp of this concept. I don't think you can, for example $F(x) = 1000x$.

If I want to be within $1000$ of $f(x)$, i.e. $\epsilon = 1000$, then $\delta$ would be $250$.

So, $f(249) = 249,000$ - which is not within $1,000$ of $x$, if $x = 1$.

Is this correct?

Thanks!

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You are correct that delta will not always be a fourth of epsilon. –  The Chaz 2.0 Apr 23 '12 at 2:24
The $\delta$ you pick depends on (i) the function $f$; (ii) the point $x_0$; and (iii) the $\epsilon$. Note that you want $f(x)$ to be within $\epsilon$ of $f(1)$, not of $1$. For this particular function, $\delta=\epsilon/4$ is not good enough. –  Arturo Magidin Apr 23 '12 at 2:25
If this were possible there would be no need for the whole $\delta$ vs. $\epsilon$ business$\ldots\$. –  Christian Blatter Apr 23 '12 at 14:30
It is a well-known canard that Freshmen believe all functions to be linear. For example, $1/(x+y) = 1/x + 1/y$ is obvious, no? –  GEdgar Jun 1 '12 at 13:55

## 1 Answer

If $f$ is linear, $f(x)=mx+b$ and you can take $\delta=\frac{\epsilon}{ |m|}$. You should plug this into the $\delta - \epsilon$ definition to see that it is true. For differentiable $f$, you can usually use something close to $\frac \epsilon {f'}$ (note that in the previous case $f'=m$) but you might need to use something smaller. This comes from the fact that the derivative gives the best local linear approximation, but higher order terms may be a problem.

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