# Thread: One more epsilon ?

1. ## One more epsilon ?

I am given

f(x) = sq root x
Xo= 1
L=1
E= 1/4
y= sq root x

and it says

find a delta > 0 that for all x 0< abs value x-Xo < alpha -> abs value f(x)-L < E

I know on the right side I will have sq root x - 1 < 1/4

I am not sure what I put on the left side however. I know what Xo is but what about the x and the alpha?

I am so confused

2. Originally Posted by KarlosK
I am given

f(x) = sq root x
Xo= 1
L=1
E= 1/4
y= sq root x

and it says

find a delta > 0 that for all x 0< abs value x-Xo < alpha -> abs value f(x)-L < E

I know on the right side I will have sq root x - 1 < 1/4

I am not sure what I put on the left side however. I know what Xo is but what about the x and the alpha?

I am so confused
You want to find a δ>0 such that :

for all ,x : if o<|x-1|< δ ,then $|\sqrt{x}-1|<\frac{1}{4}$.................................................. ............................................1

But $|\sqrt{x}-1| = \frac{|(\sqrt{x}-1)(\sqrt{x}+1)|}{|\sqrt{x}+1|}=\frac{|x-1|}{\sqrt{x}+1}$ $\leq|x-1|$.................................................. ................................................2

Since $\sqrt{x}\geq 0\Longrightarrow\sqrt{x}+1\geq 1\Longrightarrow\frac{1}{\sqrt{x}+1}\leq 1$ $\Longrightarrow\frac{|x-1|}{\sqrt{x}+1}\leq |x-1|$

So from (2) we see that if we put |x-1|<1/4 ,then $|\sqrt{x}-1|<\frac{1}{4}$.

Obviously δ = 1/4