Recent content by hoaver

Homework Statement Prove that \frac{dy}{dx}=\frac{y}{x} for \sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=10 x is not equal to y which is not equal to 0 The Attempt at a Solution Tried all the normal methods but none seem to work...anyone have any ideas?

Thats the one. Thanks =)

Not sure what you mean. The only restriction I'm guessing would be that any negative roots you get will never be a solution to an exponential expression, where the exponent is not 1.

Let y=3^x y^2-y-12=0 (y-4)(y+3)=0 y=4\ or\ y=-3 3^x=4\ or\ 3^x=-3 \log 3^x = \log 4 x= \frac{\log 4}{\log3} x \approx 1.2618595071429 For any value of x, 3^x\neq-3, this solution is extraneous (rejected).

Thanks for the help everyone. Later, I also found it to work with using this: \lim_{x \to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right)*\frac{3+\sqrt{2x+1}}{3+\sqrt{2x+1}} Simplifies to \lim_{x \to 4}\left(\frac{(2-\sqrt{x})(3+\sqrt{2x+1})}{8-2x}}\right) to \lim_{x \to...

We have not learned L'Hopital's rule or derivatives yet =(

Ah, okay, I see how to solve the 2nd one now, thanks =). Hopefully someone knows how to solve the 1st one

Homework Statement \lim_{x\to 4}\left(\frac{2-\sqrt{x}}{3-\sqrt{2x+1}}\right) \lim_{x\to 0}\left(\frac{2^2^x {-2^x}}{2^x{-1}}\right) The Attempt at a Solution I tried rationalizing but doesn't really help in evaluating these limits. If a limit cannot be factored, what...