# Thread: Find delta given epsilon

1. ## Find delta given epsilon

Could you help me with the problem?

Find delta using the definition of limits, given epsilon = 0,25

lim 1 / (2-x) = -1/3
x->5

Answer should be delta = 1
How can I get it?

Thanks.

2. Do you really need to apply a $\delta - \epsilon$ proof here?

3. Originally Posted by sodk
Could you help me with the problem?

Find delta using the definition of limits, given epsilon = 0,25

lim 1 / (2-x) = -1/3
x->5

Answer should be delta = 1
How can I get it?

Thanks.
No, the answer is not 1- but 1 is a perfectly valid answer.

The definition of "limit" says that, given any $\epsilon> 0$, there exist $\delta> 0$ such that if $|x- 5|< \delta$ then $\left|\frac{1}{2- x}- (-\frac{1}{3}\right|< \epsilon$.

Here, you are given that $\epsilon= 0,25$ so we want to have $\left|\frac{1}{2-x}+ \frac{1}{3}\right|< 0.25$

$\frac{3}{3(2-x)}+ \frac{2- x}{3(2-x)}=\frac{5- x}{3(2-x)}$

So we want $\left|\frac{5-x}{3(2-x)}\right|< .25$.

That is the same as $\left|5-x|\right|< \frac{3}{4}|x-2|$

Let's say that |5- x|< 1. Then -1< x- 5< 1 so 4< x< 6. then 2< x- 2< 4 so 2< |x-2|< 4. Certainly, $\frac{3}{4}(2)= \frac{3}{2}< \frac{3}{4}|x-2|$ then, so if we can make $|x- 5|< \frac{3}{2}$ we are done. Since we are already assuming $|x- 5|< 1$, and $1< \frac{3}{2}$, it is sufficient to make $|x- 5|< 1$. That is, we can take $\delta= 1$.

Notice that this is not saying that $\delta$ has to be equal to 1. Obviously any smaller number would work and there might be some larger value that would work. But $\delta= 1$ is sufficient.

Had I said "Let's say that |5- x|<" some other number, I would have gotten an answer other than 1, but equally valid.