Let \[f(x) = x^2 - 2x +1 \] use the epsilon-delta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\] I've come up with this: \[|( x-1)^2 - 4| < \epsilon \] \[| x-3| < \sqrt{\epsilon } \] Set delta as the squareroot of epsilon: \[| x-3| < \delta =\sqrt{\epsilon } \] How do I finish this, I get it all wrong by taking the left expresson to the second power

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Let \[f(x) = x^2 - 2x +1 \] use the epsilon-delta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\] I've come up with this: \[|( x-1)^2 - 4| < \epsilon \] \[| x-3| < \sqrt{\epsilon } \] Set delta as the squareroot of epsilon: \[| x-3| < \delta =\sqrt{\epsilon } \] How do I finish this, I get it all wrong by taking the left expresson to the second power

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

How do you have the limit is 4? You won't be able to prove what you said. That does x->2 right? So f->2^2-2(2)+1=4-4+1=0+1=1
Well, that the limit is 4 was given in the assignment and \[f(x)=x2−2x+=4 \]. where x=3 I think you got the limit wrong it's \[ x \rightarrow 3\]
Ok. I couldn't read that little number.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Thanks. So you have |x^2-2x+1-4|
E is epsilon d is delta by the way :)
I hope that helps. I will try to check back with you later. I have to go for now, but I return in like 2 hours :(
Got it! Thanks a lot, really appreciate it! :D
Great remember when showing that limit above you have to say let d=min{1, E/|x+1| where you have chosen x based off letting d<1}
Oh and wow. I can read that 3 on this computer but not that other one.

Not the answer you are looking for?

Search for more explanations.

Ask your own question