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f(x)= 1/ square root of x+2
find the f '(a)
lim h->0 (cos( 3.14 +h) + 1)/h
find the f (x) and a.
thx alot
I'm guessing you use ε on your y-axis, and δ on your x-axis (like we do over here). If so, consider this problem (same principle, I'm just using x² instead):
Show that x² --> 4 when x --> 2.
We need to show that, as soon as we've picked our ε>0, we can find a δ>0 so that |x²-4|<ε when 0<|x-2|<δ. Don't forget the 0 right there!|x²-4| = |x+2|*|x-2|
Let ε>0. If we demand |x-2|<1 (namely that 1<x<3), we get |x+2|<5. If we also demand that |x-2|<ε/5 (notice why we picked 5 below, and not any other number), we get:
|x²-4| = |x+2|*|x-2|<5*ε/5=ε
Thus, if δ is the least of the numbers 1 and ε/5, we get that |x²-4|<ε when 0<|x-2|<δ. Maybe someone could pull out a graph of the definition to make things easier...
That's all there is, I think. I don't see where you got that ε/3=δ from though?
i don't really understand this part... why we have to demand |x-2|<1...
and why the |x+2| <5 can lead the next step:|x-2|<ε/5 ? is it some kind of basic inequality properties?
i am confused... thx for help anyway
ehm, lim of x to what? to infintity? to 0, to -infinty? what?
or do you mean
in which case, wouldnt this be satisfactory
lim(x->2) 3x = 3*2 = 6, -2 = 4 ?
i mean lim of x to 2 and the function is (3x-2)
thx
prove limx->2(3x-2)=4
that means i have to prove: ∣(3x-2)-4∣<ε whenever∣x-2∣<δ
next step: ∣(3x-6)∣<ε whenever∣x-2∣<δ
next step: ∣x-2∣<ε/3whenever∣x-2∣<δ
for the next step , we let
ε/3=δ
∣(3x-2)-4∣=∣(3x-6)∣=3∣x-2∣<3δ=3(ε/3)=ε
thus ∣(3x-2)-4∣<ε whenever∣x-2∣<δ
is that correct? um... coz i 'm not really understand why we need to let ε/3=δ
thx for help
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