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A curve is defined by the equation $x^2+y^2=16(x^2-y^2)$. Find all points on the curve at which the tangent line is horizontal.

I have differentiated the equation and found $$\frac {dy}{dx} = \frac {8x-xy^2-x^3}{x^2y+y^3+8y}.$$ Then, I don't know how to solve $$8x-xy^2-x^3 = 0.$$

Also, I computed the limit shown below. The answer is 2, but the correct answer is $\ln 12^\frac13$:

$$\lim_{x\to0^+} \left(\frac133^x+\frac232^x\right)^\frac1x$$

$$=\lim_{x\to0^+} \left(\frac{\frac133^x\ln3+\frac232^x\ln2}{(\frac133^x+\frac232^x)}\right)$$ $$=\lim_{(\frac{2}{3})^x\to0^+} \left(\frac{\frac13\ln3+\frac23\frac23^x\ln2}{(\frac13+\frac23(\frac23)^x)}\right)$$

How can I solve it?

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Are you sure you copied the equation of the curve correctly? $x^2+y^2=16(x^2-y^2)$ is just a pair of intersecting lines. –  Sarastro Dec 4 '12 at 10:46
    
@hugo, it'd be a good idea to upvote any answer you find helpful, though you can choose only one as "accepted answer" –  DonAntonio Dec 4 '12 at 16:23

1 Answer 1

up vote 1 down vote accepted

$$(1)\;\;\;\;\;\;\;0=8x-xy^2-x^3=-x(x^2+y^2-8)\Longrightarrow x=0\,\,\,\vee\,\,\,x=\pm\sqrt{8-y^2}$$ $${}$$ $$(2)\;\;\;\;\;\;\;\;\;\;\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)^{1/x}=3\,\left(\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}\right)^{1/x}$$

but

$$\left(\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}\right)^{1/x}=\exp\left(\frac{\log\left(\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}\right)}{x}\right)$$

and applying L'Hospital (check this is appliable!):

$$\lim_{x\to 0^+}\frac{\log\left(\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}\right)}{x}=\lim_{x\to 0^+}\;\;\log\frac{2}{3}\cdot\frac{\left(\frac{2}{3}\right)^{x+1}}{\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}}=\frac{2}{3}\log\frac{2}{3}$$

Thus

$$\exp\left(\frac{\log\left(\frac{1}{3}+\left(\frac{2}{3}\right)^{x+1}\right)}{x}\right)\xrightarrow [x\to 0^+]{} \left(\frac{2}{3}\right)^{2/3}$$

so the limit, finally!, is:

$$3\,\left(\frac{2}{3}\right)^{2/3}$$

I can't see any $\,\ln\,$ or whatever in it...

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