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# could someone help me with my math problem im tired and i want

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could someone help me with my math problem im tired and i want to go to bed ;/
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you're a retard if you don't know how to do this
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>>17979655
If you really need help go on /sci/ and go to the stupid questions thread
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53KB, 500x465px
>>17979655
>>17979678

The answer is - 2x. Go there for the process. In short, you'll subtract out the x^2 and be left with the - 2*(x+h) term from the polynomial. Taking the limit gives you the answer
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First you gotta start with real analysis instead of pleb level "let h go to zero". I recommend starting with the axiom of completeness or some equivalent formulation, or try to construct the reals. Next, to prove limit properties, familiarize yourself with godawful delta epsilon. Finally, after being safe assured that the limit does exist, we can compute the limit by """""cancelling"""""". I suggest you go on with deriving other theorems in real analysis, such as Bolzano-Weierstrass. XXDDDDDD

HAVE FUN
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>>17979879

you sound like a sophomore math major trying hard to look smart.
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>>17979655
(x+h)^2 = x^2 + 2*x*h + h^2

lim (r(x+h) - r(x))/h = lim ((2 - (x^2 + 2*x*h + h^2)) - (2-x^2))/h = lim (-2*x*h - h^2)/h = lim (-2*x - h) = -2x
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Another easy way to look at it is
F'(x) = nx^(n-1)

So knowing that you have
R(x)= 2-x2
Which is the same as saying
R(x)= 2x^0 - x2
R'(x)= (0)2x^(0-1) - 2x^(2-1)
R'(x)= 0 - 2x
R'(x)= -2x
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>>17979879
If you are done being a dick and look at the problem you can see this anon is just starting calculus 1. Excuse him/her for not being as educated in the subject as you are with practice they will be at your level eventually.