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$$ f(x,y) = \begin{cases} \dfrac{2x^4y - 5x^2y^2+y^5}{(x^2+y^2)^2}, & \text{if $(x,y) \neq (0,0)$} \\ 0 & \text{if $(x,y) = (0,0)$} \end{cases}$$

I know this function to be discontinuous by putting $y=mx$, but i need help to prove the discontinuity using $\epsilon-\delta$ form, where i know that if a function is discontinuous then there $\exists \epsilon $ such that $\forall \delta \gt 0, |x-0|\lt \delta \Rightarrow |f(x,y) - f(0,0)| \gt \epsilon$.

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  • $\begingroup$ $y=mx $ gives nothing. $\endgroup$ Mar 17, 2018 at 16:55
  • $\begingroup$ after putting y=mx, use limits on x to get a non-zero value $\endgroup$
    – Sarkar
    Mar 18, 2018 at 8:53

1 Answer 1

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Take $\varepsilon=1$. Since, if $x\neq0$,$$f(x,x)=\frac{2x^5-5x^4+x^5}{4x^4}=\frac{3x}4-\frac54,$$for every $\delta>0$, there is a $x\in\mathbb R$ such that $\|(x,x)\|<\delta$ and $\|f(x,x)\|>1$.

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