Question

Prove that \(\lim_{x \rightarrow 0} \sqrt{4-x} =2\) using the \(\epsilon-\delta\) definition.

Answer 1

if you plug 0 in for x you get squareroot of 4 which equals 2...is this what you want?

Answer 2

Nope, I need to prove it..

Answer 3

Here's my work so far.
Let \(x_0\) =3, f(x) = \(\sqrt{4-x}\) and L=2 in the definition of limit. For any given \(epsilon\)>0, we have to find a suitable \(\delta\) >0 so that if x \(ne\) 0 and x is within distance \(\delta\) of \(x_0\) = 0, that is whenever 0< | \(\sqrt{4-x}\) | < \(\delta\), it is true that f(x) is within distance \(\epsilon\) of L=2. So, |f(x) -2} < \(\epsilon\).
Then, I don't know how to continue.

Answer 4

these are confusing but i happen to be learning it too right now wanna walk through it?

Answer 5

Sure! But I don't really understand this topic!

Answer 6

im attempting to learn too ive been working on it all day

Answer 7

ill wait for daniel

Answer 8

I'm having trouble seeing a solution, but I can walk you through the definition. So, the definition is that for every epsilon > 0 there exists a delta > 0 st. |x-c|<delta =>
|f(x)-L|<epsilon.

So, |x-c| = |x-0| = |x| < delta. You want to define delta in terms of epsilon. So, you have:

|f(x) - L| = |sqrt(4-x) - 2|

You want this to be less than epsilon. So, you somehow need to play with the inequalities until you have |sqrt(4-x) -2| < k *|x|. Then, letting delta = epsilon/k, you would have:

|sqrt(4-x) - 2| < k*|x| < k*(epsilon/k) = epsilon.

Does that make sense? The only missing step is taking |sqrt(4-x)-2| into something which is in terms of your delta.

Answer 9

1. state |x-a|
= |x-0|
=|x|

2. so if 0<|x|<δ then |sqrt(4-x) - 2|<ϵ, so this is the hard part we need to get
|sqrt(4-x) - 2|<ϵ to equal |x|<ϵ. possibly multiplying by the conjugate im stuck at this part

Answer 10

I'm sorry for late reply. Here's what I've copied during my lesson:
For any given small ϵ > 0, (so small such that our proof will work, say <1), then choose δ = ϵ. Now, whenever 0< |x-0| < δ, we have \(|\sqrt{4-x}-2| =\frac{|x|}{\sqrt{4-x}+2} \) < |x| < δ = ϵ

But... I don't understand the above.. especially why |x| suddenly appears..