Question

Will someone please help me with an epsilon delta proof

Answer 1

sure thing bud

Answer 2

A proof of a formula on limits based on the epsilon-delta definition. An example is the following proof that every linear function f(x)=ax+b (a,b in R,a!=0) is continuous at every point x_0. The claim to be shown is that for every epsilon>0 there is a delta>0 such that whenever |x-x_0|<delta, then |f(x)-f(x_0)|<epsilon. Now, since

|f(x)-f(x_0)| = |ax+b-(ax_0+b)|]
(1)
= |ax-ax_0|
(2)
= |a|·|x-x_0|,
(3)
it is clear that

|x-x_0|<epsilon/(|a|) implies |f(x)-f(x_0)|<|a|·epsilon/(|a|)=epsilon.

Answer 3

Sorry I have like x^2 in my problem and I have no idea how to do it

Answer 4

Its the limit of x^2+x as x approaches 3 is 12

Answer 5

oh i see

Answer 6

Yeah

Answer 7

I can do the first couple steps than I get lost

Answer 8

\[lim_{x\rightarrow 3}x^2+x =12\]
right?

Answer 9

You have to go forward on wrap paper, then go backward on the proof . Like this:
On wrap paper, you work for value of \(\epsilon,\delta\), and the relationship between them.
On the paper you will hand on, you go from the result.

Answer 10

On wrap paper:
By definition,
\(0<|x-3| <\delta\) \(\rightarrow 0<|x|<\delta +3\rightarrow 0<|x+4|<\delta +7\)

\(0<|x^2+x-12|<\epsilon\)

so, \(0<|x-3||x+4|<\delta *(\delta +7)=\epsilon\)

Now, solve for relationship between \(\delta ~and ~\epsilon\)

the far right: d^2+7d -e=0, solve this quadratic to get
\(\delta =\dfrac{-7\pm\sqrt {49+4\epsilon}}{2}\)

But \(\delta >0\), hence only \(\delta = \dfrac{-7+\sqrt{49+4\epsilon}}{2}\) works.

However, you need this \(\delta >0\), that force \(-7+\sqrt{49+4\epsilon}>0\)

or \(49+4\epsilon>49\), or \(\epsilon >0\)

Answer 11

Now, on handed paper, you write
\(\forall \epsilon >0, \exists \delta =\dfrac{-7\pm\sqrt{49+4\epsilon}}{2}\) such that, when \(0<|x-3|<\delta\), \(0<|x^2+x-12|< \epsilon\)
Then jot down the proof from the wrap paper in
good luck