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

Calculus1
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Try combining top fractions. And get rid of the compound fraction. Unless you already know derivatives then we can actually skip this part.
so a common denominator for the top fractions? do i expand the (x+h)^2
You will need to.

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Other answers:

so x^2 + 2xh + h^2, what would be the common denominator
x^2(x+h)^2
|dw:1379983226588:dw|
|dw:1379983241382:dw|
2xh + h^2/ x^2 (x+h)^2 x (1/h)
can i factor out a h from the numerator?
so i factored out an h from the numerator, than cancelled the h from 1/h
What happen to your negative in from 2xh?
h/h=1 yep yep
i have lim h--0 2x+h/ x^2(x+h)^2
so now i can "sub" in 0 into h, but what value goes for x?
Well yeah but still you are missing a negative on top
only h is going somewhere x stays
It said what happens as h goes to zero (it said nothing for x)
Did you figure out what negative you are missing?
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
because I factored out an h, but im confused as to my next step
h goes to 0
is my answer suppose to be just a number? or will it have a value and x
Your answer will be a function of x.
-2x-0/x^2 (x+0) ^2 ?
(-2x-0)/(x^2(x+0)^2) yes Simplify.
-2/x?
not rlly syre how to simplify the denominator
\(\cfrac{(-2x-0)}{(x^2)(x+0)^2}\) (Just for ease - wrote the LaTeX)
-2x-0=-2x x^2(x+0)^2=x^2(x)^2 (since x+0=x) Use law of exponents.
Recall one of the laws of exponents a^m * a^n=a^(m+n)
ok so my answer is -2/x^3
yep
thank you :)
But the denominator should be x^4 @myin
-2x/x^4=-2/x^3
I have another question: find the limit if it exists limx-- -2 2-|x|/ 2+x
thats lim x approaches -2
can I just sub in -2 for x? so that makes it 0/0
Oh sorry @myininaya - didn't saw that :) Thanks !
\[\lim_{x \rightarrow -2} \frac{2-|x|}{2+x}\] ?
yes
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
That is your key ingredient
okay

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