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myininayaBest ResponseYou've already chosen the best response.1
Try combining top fractions. And get rid of the compound fraction. Unless you already know derivatives then we can actually skip this part.
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
so a common denominator for the top fractions? do i expand the (x+h)^2
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
so x^2 + 2xh + h^2, what would be the common denominator
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
dw:1379983226588:dw
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
dw:1379983241382:dw
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
2xh + h^2/ x^2 (x+h)^2 x (1/h)
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
can i factor out a h from the numerator?
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
so i factored out an h from the numerator, than cancelled the h from 1/h
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
What happen to your negative in from 2xh?
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
i have lim h0 2x+h/ x^2(x+h)^2
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
so now i can "sub" in 0 into h, but what value goes for x?
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
Well yeah but still you are missing a negative on top
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
only h is going somewhere x stays
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
It said what happens as h goes to zero (it said nothing for x)
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
Did you figure out what negative you are missing?
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
ok i think i see my mistake so i went thru the stages again and right now I have lim h0 2xh/x^2 (x+h)^2
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
because I factored out an h, but im confused as to my next step
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
is my answer suppose to be just a number? or will it have a value and x
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
Your answer will be a function of x.
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
(2x0)/(x^2(x+0)^2) yes Simplify.
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
not rlly syre how to simplify the denominator
 7 months ago

mathsloverBest ResponseYou've already chosen the best response.0
\(\cfrac{(2x0)}{(x^2)(x+0)^2}\) (Just for ease  wrote the LaTeX)
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
2x0=2x x^2(x+0)^2=x^2(x)^2 (since x+0=x) Use law of exponents.
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
Recall one of the laws of exponents a^m * a^n=a^(m+n)
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
ok so my answer is 2/x^3
 7 months ago

mathsloverBest ResponseYou've already chosen the best response.0
But the denominator should be x^4 @myin
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
I have another question: find the limit if it exists limx 2 2x/ 2+x
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
thats lim x approaches 2
 7 months ago

muffinBest ResponseYou've already chosen the best response.0
can I just sub in 2 for x? so that makes it 0/0
 7 months ago

mathsloverBest ResponseYou've already chosen the best response.0
Oh sorry @myininaya  didn't saw that :) Thanks !
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
\[\lim_{x \rightarrow 2} \frac{2x}{2+x}\] ?
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
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
 7 months ago

myininayaBest ResponseYou've already chosen the best response.1
That is your key ingredient
 7 months ago
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