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$$ \lim\limits_{(x,y) \to (0,0)} \frac{2x^2y^3}{x^4+y^6} $$

My reasoning after reading the textbook: Direct substitution wouldn't work since it would lead to the indeterminate form $0/0$. We can examine the values of $f$ along parabolic curves that end at $(0,0)$. Along $y=kx^2, x\neq0$, the function has the value

$$ \frac{2k^3x^4}{1+k^6x^8}. $$

So, if $(x,y)$ approaches $(0,0)$ along $y=x^2$, then $k = 1$, and the limit is $\frac{2x^4}{1+x^8}$. If $(x,y)$ approaches $(0,0)$ along the $x$-axis, then $k=0$, and the limit is $0$.

Is this correct and/or is there another way to go about this? Thanks.

Just_A_User
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    As $(x,y)$ approaches $(0,0)$, $\frac{2x^4}{1+x^8}$ approaches $0$ as well. – DMcMor Jul 13 '20 at 02:18
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    Suggestion: try $y\propto x^{2/3}$. Why? By the AM-GM inequality, $2x^2y^3 \leq x^4+y^6$, with equality iff $x^2 = y^3$. – Clement C. Jul 13 '20 at 02:22

1 Answers1

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As DMcMor observed, unfortunately your reasoning is incorrect as $$\frac{2x^4}{1 + x^8} \to 0$$ as $x \to 0$. But the idea is correct: you want to find two paths through $(0, 0)$ whose limits are different. You might consider the paths $y = x^{2/3}$ and $y = x$, for example. Usually, in problems where we want to show the limit doesn't exist, the strategy for picking two paths is to first find a path where the numerator's degree is higher than the denominator's degree (happening here with $y = x$) --- in which case the limit becomes zero --- and then find a second path through which the denominator and numerator have the same degree (happening here with $y = x^{2/3}$ --- in which case the limit will be nonzero.

paulinho
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  • I can't believe I didn't see that! Anyways, thanks a lot for this -- really helpful. –  Jul 13 '20 at 23:43