A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\]
I've come up with this:
\[( x1)^2  4 < \epsilon \]
\[ x3 < \sqrt{\epsilon } \]
Set delta as the squareroot of epsilon:
\[ x3 < \delta =\sqrt{\epsilon } \]
How do I finish this, I get it all wrong by taking the left expresson to the second power
anonymous
 4 years ago
Let \[f(x) = x^2  2x +1 \] use the epsilondelta definition to show that: \[\lim_{x \rightarrow 3} f(x) = 4\] I've come up with this: \[( x1)^2  4 < \epsilon \] \[ x3 < \sqrt{\epsilon } \] Set delta as the squareroot of epsilon: \[ x3 < \delta =\sqrt{\epsilon } \] How do I finish this, I get it all wrong by taking the left expresson to the second power

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It can be done something like this \[f(x)= x^22x+1 \] and \[\lim_{x\rightarrow3}x^22x+1=4 \] first set $$\epsilon >0$$ Now find a $$\delta > 0$$ That relies on $$\epsilon$$ First we note that $$ \left (x^22x+1)4\right < \epsilon$$ can be stated like $$ \left x^22x3 \right < \epsilon $$ and then $$ \left x+1\right \left x3\right < \epsilon $$ Now we are actually looking for a delta, to do this we need to make use of the x+1 and get some sort of value for it so we can rewrite the above with the x3 term. To do this we first make the statement that $$ \delta \leq 1$$ this is always useful for these types of limit proofs because, A it gives us a number to work with, and B its a fair assumption as less than or equal to 1 is probably the GREATEST you would want in finding values for delta. Ok, so now lets get some values for x+1, given that $$  x3 < \delta \leq 1$$ we can write $$ x3 < 1 $$ this implies $$ 1 < x3 < 1 $$ which also implies $$ 2 < x <4$$ Ok, now we are in a great position, we have some values to use to change the x+1 to a number using the above. $$ 3 <  x+1 < 5 $$ Now we can continute with our epsilon $$ x+1 x3 < x3 \cdot 5 < \epsilon$$ Giving us $$ x3 < \frac{\epsilon}{5} $$ Which we can use for our delta ( providing that this value is < 1, if not then choose 1) $$ \delta = \frac{\epsilon}{5}$$
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.