Question

f(x)=2(x^(1/2))-Ax. A>0. find the lim f'(x) as x goes to 0. I'm thinking that the answer is infinity, but I was just wondering if someone could please check it

Answer 1

\[f(x)=2\sqrt{x}-Ax\] right?

Answer 2

so
\[f'(x)=\frac{1}{\sqrt{x}}-A\]

Answer 3

yeah, so then the derivative is x^(-1/2)+A

Answer 4

there is no limit as \(x\to 0\) or i guess you could say \(\infty\)

Answer 5

oh, i meant -A

Answer 6

ok, thanks

Answer 7

no matter, still no limit

Answer 8

I also have another question, could you check and see if the second derivative of the original function is -1/(2x(x^1/2))

Answer 9

sorry i don't know how to do the square root

Answer 10

\sqrt{a}

Answer 11

but your answer is correct in any case

Answer 12

ok, thank you!

Answer 13

\[-\frac{1}{2x^{\frac{3}{2}}}\] is one version, or what you wrote
yw