## anonymous 5 years ago Factoring Advice: Goal is to eliminate the x's. [(x+y)^2 - x^2] / y

1. anonymous

it's equal to 2x+y , eliminate the x's ?

2. anonymous

oh that's different than the first one

3. anonymous

I think there is a way to do it, but I am not sure how I can get them to all cancel out.

4. anonymous

$((x+y)2−y2)/y$ goal is to eliminate the x's and be left with only y's to deal with...

5. anonymous

Thoughts? I have tried the usual things I can think of, but unless I made an error I am not catching, I am stumped.

6. anonymous

BRB

7. anonymous

is this part of a LIM problem ?

8. anonymous

Yes, how on earth did you guess?

9. anonymous

I mean are you looking for: $\lim_{y \rightarrow 0} \frac{(x+y)^2-x^2}{y}$

10. anonymous

Uh, indeed I am...

11. anonymous

Can you help me get unstuck? I really want to learn how to do this...

12. anonymous

I have it down to y + 2x and I don't know if I can use x in my lim answer, I've never seen a variable in a lim.

13. anonymous

$\lim_{y \rightarrow 0} \frac{\cancel{x^2}+2xy+y^2-\cancel{x^2}}{y}=\lim_{y \rightarrow 0}2x+y=2x$

14. anonymous

Is it acceptable to have a variable in the lim solution?

15. anonymous

yes, coz it's basically the derivative: $(x^2)'=2x$

16. anonymous

Or is this DNE?!?

17. anonymous

the final answer is 2x which is the derivative of x^2

18. anonymous

I don't know enough to understand that in the context of this problem. I understand what a first derivative is, and how you showed that, but I do not understand it in relation to my first "evaluate lim's" homework assignment.

19. anonymous

how was the problem worded ?

20. anonymous

So the lim is really 2x?!?

21. anonymous

The lim stmt you wrote, preceded by the word "Evaluate:"

22. anonymous

yes, so the answer would be 2x

23. anonymous

We haven't covered derivatives yet - not until tomorrow. So this is kinda tricky!

24. anonymous

this is how they introduce you to derivatives

25. anonymous

you don't have to know what a derivative is to solve the LIM problem. But in effect, they will reveal that what you were doing was calculating the derivative.

26. anonymous

Have you seen Gilbert Strang's high school calculus course on MIT OCW, called "The Big Picture of Calculus?" THAT's how to teach calculus.

27. anonymous

Are you a student or faculty?

28. anonymous

that formula is basically calculating rise over run (where they make the run smaller and smaller)

29. anonymous

Ahhhh

30. anonymous

neither. I studied math about 20 years ago.

31. anonymous

What do you do now that allows you to stay fresh on it? I am 44, BTW.

32. anonymous

we are very close in age :) I just started solving problems on openstudy to keep the mind from rusting too much.

33. anonymous

Well, thank you for your help. I am now a fan.

34. anonymous

BTW, I started going over the MIT OCW single variable calculus - but then got sucked in over here in the math group.

35. anonymous

I want to build up a community on MIT OCW SCHOLAR SVC.

36. anonymous

I am trying to login there every night at 8 PM. I missed it tonight, the first time others were there, too.

37. anonymous

So, if you are around at 8PM ET, I will try to be there in the MIT OCW SVC (NOT the 18.01 - the SCHOLAR version. It is a new and improved course, totally self contained.

38. anonymous

Thanks. Over and out on this case. I had solved it and didn't realize I was done.

39. anonymous

My schedule in the evenings is not fixed, juggling around kids & work . And I didn't continue past the first OCW lecture - btw I didn't know they now have the new and improved course.

40. anonymous

It is wonderful. I hope you find time to give it a try during your "me time."

41. anonymous

I'm going to resume my homework. I have about a zillion more to go...

42. anonymous

OK :)