f(x)=2(x^(1/2))-Ax. A0. 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

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

oh, i meant -A

anonymous · Answer

ok, thanks

anonymous · Answer

no matter, still no limit

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

\sqrt{a}

anonymous · Answer

but your answer is correct in any case

anonymous · Answer

ok, thank you!

anonymous · Answer

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