anonymous
  • anonymous
Evaluate the limit, if it exists lim h-0 ((1/(x+h)^2 - 1/x^2 ))/ h
Calculus1
katieb
  • katieb
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myininaya
  • 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
  • anonymous
so a common denominator for the top fractions? do i expand the (x+h)^2
myininaya
  • myininaya
You will need to.

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

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