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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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    @hugo, it'd be a good idea to upvote any answer you find helpful, though you can choose only one as "accepted answer"2012-12-04

1 Answers 1

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

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...