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