Evaluate the limit, if it exists
lim h-0 ((1/(x+h)^2 - 1/x^2 ))/ h

- anonymous

Evaluate the limit, if it exists
lim h-0 ((1/(x+h)^2 - 1/x^2 ))/ h

- schrodinger

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- myininaya

Try combining top fractions. And get rid of the compound fraction.
Unless you already know derivatives then we can actually skip this part.

- anonymous

so a common denominator for the top fractions? do i expand the (x+h)^2

- myininaya

You will need to.

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## More answers

- anonymous

so x^2 + 2xh + h^2, what would be the common denominator

- myininaya

x^2(x+h)^2

- myininaya

|dw:1379983226588:dw|

- myininaya

|dw:1379983241382:dw|

- anonymous

2xh + h^2/ x^2 (x+h)^2 x (1/h)

- anonymous

can i factor out a h from the numerator?

- anonymous

so i factored out an h from the numerator, than cancelled the h from 1/h

- myininaya

What happen to your negative in from 2xh?

- myininaya

h/h=1 yep yep

- anonymous

i have lim h--0 2x+h/ x^2(x+h)^2

- anonymous

so now i can "sub" in 0 into h, but what value goes for x?

- myininaya

Well yeah but still you are missing a negative on top

- myininaya

only h is going somewhere
x stays

- myininaya

It said what happens as h goes to zero (it said nothing for x)

- myininaya

Did you figure out what negative you are missing?

- anonymous

ok i think i see my mistake so i went thru the stages again and right now I have lim h--0 -2x-h/x^2 (x+h)^2

- anonymous

because I factored out an h, but im confused as to my next step

- myininaya

h goes to 0

- anonymous

is my answer suppose to be just a number? or will it have a value and x

- myininaya

Your answer will be a function of x.

- anonymous

-2x-0/x^2 (x+0) ^2 ?

- myininaya

(-2x-0)/(x^2(x+0)^2)
yes
Simplify.

- anonymous

-2/x?

- anonymous

not rlly syre how to simplify the denominator

- mathslover

\(\cfrac{(-2x-0)}{(x^2)(x+0)^2}\)
(Just for ease - wrote the LaTeX)

- myininaya

-2x-0=-2x
x^2(x+0)^2=x^2(x)^2 (since x+0=x)
Use law of exponents.

- myininaya

Recall one of the laws of exponents a^m * a^n=a^(m+n)

- anonymous

ok so my answer is -2/x^3

- myininaya

yep

- anonymous

thank you :)

- mathslover

But the denominator should be x^4 @myin

- myininaya

-2x/x^4=-2/x^3

- anonymous

I have another question: find the limit if it exists limx-- -2 2-|x|/ 2+x

- anonymous

thats lim x approaches -2

- anonymous

can I just sub in -2 for x? so that makes it 0/0

- mathslover

Oh sorry @myininaya - didn't saw that :)
Thanks !

- myininaya

\[\lim_{x \rightarrow -2} \frac{2-|x|}{2+x}\] ?

- anonymous

yes

- myininaya

So we can't just simply plug in -2 because our denominator will be 0.
Is there someway to make the function continuous at x=-2 so we can just plug in
Well |x|=x if x>=0 and -x if x<0
Since x approaches -2 then x is negative since we are looking at what surrounds a negative number
so we have |x|=-x

- myininaya

That is your key ingredient

- anonymous

okay

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