# Basic epsilon and delta proofs, limits, quick questions.

1. Dec 17, 2013

### binbagsss

I am trying to check whether lim h→0 (R(h)/||h||) =0 or not.
I am working in ℝ2.
h=h1e1+h2e2**
=> ||h||=(h1^2+h2^2)^1/2

I am using the definition that (R(h)/||h||)<ε * whenever 0<|h|<δ for all h.

Example 1
(R(h)/||h||)=h1h2/(h1^2+h2^2)^3/2
I can see that the denominator dominates, so expect the limit not to exist, so in order to proove this explicitly/epsilon-delta method I look for a counter example.

I consider h1=h2. In which case I get:
(R(h)/||h||) = 1/[2^(3/2).||h||].

My problem... My book then says 'no matter how small δ is, there are therefore points with (R(h)/||h||)>ε, for every ε.'

I am don't understand this, first of all , because the expression involves 1/||h||, I thought we would be looking to increase δ to attain *<ε.
Secondly, why cant you simply take δ=1/[ε.2^(3/2)]?

Question 2
On concepts of the definition, I am confused as to whether δ can be a function of h1 and h2,as defined by **, as well as ε. I know that the definition must hold for all h, so intuitively no, but aren't h1 and h2 completely arbitary?

Many thanks for anyone who can shed some light on this, greatly appreciated :)

2. Dec 17, 2013

### jbunniii

As $h$ approaches zero, $\|h\|$ also approaches zero, so $1/\|h\|$ becomes arbitrarily large: it certainly does not approach zero.

To state this more explicitly, choose any $\epsilon > 0$. Now $1/\|h\| > \epsilon$ whenever $\|h\| < 1/\epsilon$. This is exactly the opposite of what would happen if the limit was zero!

No, the $\delta$ can depend on $\epsilon$ but not on $h_1$ and $h_2$. Given $\epsilon$, you need to find a $\delta$ such that $|R(h)/\|h\|| < \epsilon$ for all $h_1$ and $h_2$ that satisfy $0 < \sqrt{h_1^2 + h_2^2} < \delta$.