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

- anonymous

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

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- anonymous

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

- anonymous

oh that's different than the first one

- anonymous

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

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## More answers

- anonymous

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

- 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.

- anonymous

BRB

- anonymous

is this part of a LIM problem ?

- anonymous

Yes, how on earth did you guess?

- anonymous

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

- anonymous

Uh, indeed I am...

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Or is this DNE?!?

- anonymous

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

- 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.

- anonymous

how was the problem worded ?

- anonymous

So the lim is really 2x?!?

- anonymous

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

- anonymous

yes, so the answer would be 2x

- anonymous

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

- anonymous

this is how they introduce you to derivatives

- 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.

- 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.

- anonymous

Are you a student or faculty?

- anonymous

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

- anonymous

Ahhhh

- anonymous

neither. I studied math about 20 years ago.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- 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.

- anonymous

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

- 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.

- anonymous

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

- anonymous

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

- anonymous

OK :)

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