Self studier here. I was trying to understand the epsilon-delta

>>8378172
x^2+1
let x= 5+d
(5+d)^2+1 = 26+10d+d^2

for 10d+d^2 < e
let d = min{1, e}/11

>>8378184
I don't understand why you let d=min(1,e)/11

Here's a new proof based on your insight I wrote which I think is right
Proof:
Let e>0
Suppose x = 5+d for some d>0
Suppose abs(x^2+1-26)<e

then, abs((5+d)^2-25)<e
=abs(d^2+10d+25-25)<e
=abs(d^2+10d)<e

Note
abs(d^2+10d)>=abs(d^2)-abs(10d)
Since d>0
=> abs(d^2)-abs(10d)=d^2-10d

then
d^2-10d<e
=>(d-5)^2<e+25
=d-5<sqrt(e+25)
=d<sqrt(e+25)+5

QED

>>8378172
>Suppose abs(x^2+1-26)<e
>Is this right?
No, on the chance this thread isn't bait, what you assumed is what you're supposed to prove.

Structure of an epsilon-delta limit proof:
>To prove: [math]\lim_{x\to a} f(x)=L[/math]
>For all [math]\epsilon>0[/math]
>Prove there exists some constant [math]\delta>0[/math] such that
>[math]|x-a|<\delta \Rightarrow |f(x)-L|<\epsilon[/math]

>>8378244
I think you're the one misunderstanding. I wrote suppose abs(x^2+1-26)<e
because abs(f(x)-L)=abs([x^2+1]-26)

Assume your reply isn't bait, that's why my proof was set up that way

>>8378172
Start with sequences and their limits. Once you have mastered this, then you move in to functions from the real line to itself

>>8378250
You don't prove a proposition backwards by assuming the conclusion.

I'll guide you through the proof outlined here >>8378244

You want to prove [math]\lim_{x\to 5}(x^2+1)=26[/math],
so [math]a=5[/math] and [math]L=26[/math]

Given [math]\epsilon>0, |x-5|<\delta[/math]

[math]|f(x)-L|=|(x^2+1)-26|=|x^2-25|=|x+5||x-5|<|x+5|\delta[/math]

Here we need some way to find an upper bound for the term [math]|x+5|[/math]

If we choose [math]\delta\leq 1[/math] (which we are allowed to freely pick), then

[math]|x-5|\leq 1[/math]

[math]\Rightarrow -1 \leq x-5 \leq 1[/math]

[math]\Rightarrow 9 \leq x+5 \leq 11[/math]

[math]\Rightarrow |x+5|<11[/math]

Now that we've found an upper bound:

[math]|f(x)-L|<|x+5|\delta<11\delta = \epsilon,[/math]
if we choose [math]\delta=\frac{\epsilon}{11}[/math]

Remember earlier we also required that [math]\delta\leq 1[/math],

thus to complete the proof we choose [math]\delta=\min\{1,\frac{\epsilon}{11}\}[/math]

>>8378250
no my guy. you're supposed to prove that if x is in some interval (x-d,x+d) then the function f(x) must be in some interval (L-e,L+e)

>>8378250
>I wrote suppose abs(x^2+1-26)<e
You cannot assume that [math]|(x^2+1)-26|<\epsilon[/math], you have to PROVE that this is the case.

Furthermore, your choice of [math]\delta[/math] must be a positive real number independent of x.

>>8378280
Correction: That is supposed to be [math]|x+5|\leq 11[/math]
and [math]|x+5|\delta \leq 11\delta[/math]

>>8378172
ask your professor faggot