For the limit below illustrate the definition by finding the largest possible values of δ that
correspond to ε = 0.2 and ε = 0.1

lim x-2 (x^3-4x+1)=1

if ε = 0.2 the largest possible value for δ is:

if ε = 0.1 the largest possible value for δ is:

You don't have to solve the problem, just walking me through how to do the problem step by step would help immensely. Thanks!

Question

For the limit below illustrate the definition by finding the largest possible values of δ that 
correspond to ε = 0.2 and ε = 0.1

lim x->2 (x^3-4x+1)=1 

if ε = 0.2 the largest possible value for  δ  is: 


if ε = 0.1 the largest possible value for  δ  is: 


You don't have to solve the problem, just walking me through how to do the problem step by step would help immensely. Thanks!

freckles · Answer

Did you try using the epsilon delta proof:

freckles · Answer

definition*

freckles · Answer

$$\lim_{x \rightarrow c}f(x)=L$$ says: $$\text{ For all } \epsilon>0 \text{ there exists } \delta >0 \text{ such that} \ \text{ for all } \text{ } x \text{ } 0<|x-c|<\delta => |f(x)-L|<\epsilon$$ So if we chose epsilon to be 0.01 and f(x)=x^2-x and c=1. Then L=0. So we have $$ \text{ there exists } \delta >0 \text{ such taht } \ \text{ for all } x \text{ } 0<|x-1|<\delta \ |f(x)-L|=|x^2-x-0|<0.01$$ .... Or try to look at this: Solve both of the equations for x: $$f(x)=L+\epsilon$$ $$f(x)=L-\epsilon$$ Then choose the x that is closest to x=1. Then plug that x into |x-c|

freckles · Answer

So we have 
$$x^2-x=.01$$
x=1.009
x=-0.00990195

$$x^2-x=-.01$$
x=.98989898
x=.0101021


The x that is closest to our c=1 is....
x=1.009
So we have $$0<|x-1|<\delta$$
So plug in x=1.009
|1.009-1|=|.009|=.009
We will choose that number to be our delta...


We still should be able to close by letting delta be between 0 and 1/2
So
we have 0<|x-1|<delta
if we choose delta to be between 0 and 1/2 we have
|x-1|<1/2
-1/2<x-1<1/2
1/2<x<3/2
So we also have
$$ |f(x)-L|=|x^2-x-0|
=|x|*|x-1|<\epsilon$$
So This means $$|x-1|<\frac{\epsilon}{|x|}$$

x is between 1/2 and 3/2 
So $$|x-1|<\frac{ 2 \epsilon}{3}$$
$$|x-1|<\frac{2}{3}(.01)=\frac{1}{150}$$

but as you see the delta from before was larger than this one
so that method above might be better

anonymous · Answer

A slightly different (but fundamentally similar) approach would be to prove the limit using the definition, then plugging in the given $\epsilon$.
$$\begin{align*}\left|x^3-4x+1-1\right|&=\left|x^3-4x\right|\\
&=|x||x+2||x-2|\\
&<3\times5~|x-2|&\text{(agree to set }\delta\le1)\\
&<\epsilon
\end{align*}$$
which would mean $\delta=\min\left\{1,\dfrac{\epsilon}{15}\right\}$.

For $\epsilon=0.2$, you have $\delta=\dfrac{0.2}{15}=\dfrac{2}{150}$, and for $\epsilon=0.1$, $\delta=\dfrac{1}{150}$ (as freckles has shown).

freckles · Answer

your delta is 2/150=1/75

freckles · Answer

oops nevermind

freckles · Answer

looking at the other epsilon and didn't look anything in between

freckles · Answer

I wonder if that method gives us the largest delta 
He is looking for largest delta

freckles · Answer

I think you have to find when f(x)=L+epsilon and find when f(x)=L-epsilon
and then get the x from there that is closest to your c (then number x is approaching)
Then take that c and evaluate |x-c| to find the largest possible delta

freckles · Answer

could be wrong though