Solving an Equation with h=0: Is 2-2x the Answer?

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so i was working on this equation , and in the end i came to
Lim H>0 2-h+2xh
so now i am supposed to treat H as 0
so shouldn't it be 2-(0)+2x(0) = 2 ?
i know the answer is 2-2x , as if we just removed the H , but isn't multiplying anything be zero make it a zero itself ? * 2x*0 = 0 ?
 
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Hey B4ssHunter.

Is there meant to be a h in the denominator? Is this is some sort of differentiation problem via first principles?
 
how can there be an h in the denominator. that would make it undefined
 
B4ssHunter said:
so i was working on this equation , and in the end i came to
Lim H>0 2-h+2xh
so now i am supposed to treat H as 0
so shouldn't it be 2-(0)+2x(0) = 2 ?
i know the answer is 2-2x , as if we just removed the H , but isn't multiplying anything be zero make it a zero itself ? * 2x*0 = 0 ?
IF lim(h->0) 2- h+ 2xh is, in fact, correct, then the limit is 2, not 2- 2x.

So if you you say "I know the answer is 2- 2x", then 2- h+ 2xh must be incorrect. Please tell us what the entire problem is.
 
oh i apologize , i made a sign mistake .. i am really really sorry :S
 

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