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I need to prove that

$$\lim \limits_{(x,y)\to(0,0)} \frac{5y^2\cos^2x}{x^4 + y^4}$$

does not exist.

Please tell me why this false positive argument is flawed.

However, if I put this into a delta epsilon limit, I obtain that:

$$\frac{\left|5y^4\cos^2x - (x^4+y^4)L\right|}{\epsilon} < x^4 + y^4 \qquad(1)$$

Then we need to find $\delta$ such that $\sqrt{x^2 + y^2} < \delta$ for any given $\epsilon$. Well because for $x^4 + y^4$ near zero $x^4 + y^4 \le x^2 + y^2$, by the monotonicity of $\sqrt{x}$, $$\sqrt{x^4 + y^4} \le \sqrt{x^2 + y^2}$$ Just use any $$\delta < \sqrt{\frac{\left|5y^4\cos^2x - (x^4+y^4)L\right|}{\epsilon}} \le \sqrt{x^2+y^2}$$

What about this is false?

Is it because there exists $x^4 + y^4 > x^2 + y^2$ so the limit doesn't always hold for any given $\epsilon$?

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  • $\begingroup$ Subs $$x=\frac{1}{n},y=\frac{\sqrt{n}}{n}$$ in your $$f(x,y)$$ and the Limit for $n$ tends to infinity doesn't exist. $\endgroup$ – Dr. Sonnhard Graubner Mar 23 '17 at 15:31
  • $\begingroup$ The "positive" argument is flawed because it makes no sense at all. You have this symbol $L$ that mysteriously shows up in equation (1), which you somehow "obtained" (implying you think it's true, though you give no justification), and then in order to "find $\delta$ such that $\sqrt{x^2+y^2}<\delta$ for any given $\epsilon$" (whatever that means!), you set $\delta \leq \sqrt{x^2+y^2},$ in direct contradiction of the thing you said you needed to do. $\endgroup$ – David K Mar 23 '17 at 15:47

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