When h goes to zero (it appears by way of your syntax that the whole deal is being divided by h, if that is true, as h goes to zero in the denominator, the whole thing goes to infinity.
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Quote from minno726 »
Now, if you'll excuse me, I have to make everything symmetrical.
If all that **** in the numerator does not contain an 'h', and the only thing in the denominator is 'h', then the whole thing goes to infinity as h goes to 0. Think of it as 1 being divided by .00000000000000001, that makes that number get reallly really big. Essentially that is what happening, only envision h as being .0000000000000000000000000000000000000000000000000000000000001, making that whole quantity go to infinity basically.
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Quote from minno726 »
Now, if you'll excuse me, I have to make everything symmetrical.
The numerator then becomes 3(x+h) - 3x = 3h. Cancel with the h in the denominator (since h is approaching 0 and not equal to 0 this is legitimate) and you get:
f '(x)=??
of
Limit as h->0 of f(x)= 3x^(1/2)
anyone? im so ****ing confused right now
for reference the equation of f '(x)= [f(x+h)-f(x)]/h
If x is approaching zero in this equation, 3*0^(1/2) is still 3*0 which would be zero.
([3(x+h)]^1/2- 3x^1/2)/h
but idk where im going wrong :/
When h goes to zero (it appears by way of your syntax that the whole deal is being divided by h, if that is true, as h goes to zero in the denominator, the whole thing goes to infinity.
http://anarchyinyourhead.com/
http://www.strike-the-root.com/
http://mises.org/
or f(x)= the square root of 3x
f '(x)=???
f'(x) would be -- ((1/2)*(3x)^(-1/2))*x. so actually you multiply the whole deal by x according to the chain rule
I think he wants the steps used using the limit method. I'm tearing my hair out trying this algebra I haven't needed to use in years.
And in't it (3^0.5)/(2*(x^0.5))
found it using power rule and L'Hospitales rule, but still can't get the damn algebra down.
http://anarchyinyourhead.com/
http://www.strike-the-root.com/
http://mises.org/
lim h -> 0: ([3(x+h)]^0.5 - (3x)^0.5)/h * ([3(x+h)]^0.5 + (3x)^0.5)/([3(x+h)]^0.5 + (3x)^0.5)
The numerator then becomes 3(x+h) - 3x = 3h. Cancel with the h in the denominator (since h is approaching 0 and not equal to 0 this is legitimate) and you get:
lim h -> 0: 3 * 1/(([3(x+h)]^0.5 + (3x)^0.5) = 3 * 1/((3x)^0.5 + (3x)^0.5) = 3/2 (3x)^-0.5
(you stand above us all)
Agreed, that was impressive. Mighta been able to do that sober, and with my calc book as a guide haha.