Question

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

Answer 1

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

Answer 2

oh that's different than the first one

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

BRB

Answer 7

is this part of a LIM problem ?

Answer 8

Yes, how on earth did you guess?

Answer 9

I mean are you looking for:

\[\lim_{y \rightarrow 0} \frac{(x+y)^2-x^2}{y}\]

Answer 10

Uh, indeed I am...

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

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

Answer 15

yes, coz it's basically the derivative:

\[(x^2)'=2x\]

Answer 16

Or is this DNE?!?

Answer 17

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

Answer 18

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.

Answer 19

how was the problem worded ?

Answer 20

So the lim is really 2x?!?

Answer 21

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

Answer 22

yes, so the answer would be 2x

Answer 23

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

Answer 24

this is how they introduce you to derivatives

Answer 25

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.

Answer 26

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.

Answer 27

Are you a student or faculty?

Answer 28

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

Answer 29

Ahhhh

Answer 30

neither. I studied math about 20 years ago.

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

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

Answer 39

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.

Answer 40

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

Answer 41

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

Answer 42

OK :)