Continuity of piecewise function of two variables

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SUMMARY

The piecewise function defined as $f(x, y) = 0$ for $y \leq 0$ or $y \geq x^4$, and $f(x, y) = 1$ for $0 < y < x^4$, is proven to be discontinuous at the point (0, 0) and along the curves $y = 0$ and $y = x^4$. The discontinuity at (0, 0) arises because for any $\varepsilon > 0$, it is impossible to find a $\delta$ such that all points within a distance $\delta$ from (0, 0) yield $|f(x, y)| < \varepsilon$. Additionally, along the curves, points exist where $f$ takes both values 0 and 1, confirming the discontinuity along these paths.

PREREQUISITES
  • Understanding of piecewise functions
  • Familiarity with continuity and discontinuity definitions in multivariable calculus
  • Knowledge of limits and $\varepsilon$-$\delta$ proofs
  • Basic graphing skills for visualizing curves and functions
NEXT STEPS
  • Study the definition of continuity for multivariable functions
  • Explore examples of piecewise functions and their continuity properties
  • Learn about the implications of discontinuity on the behavior of functions
  • Investigate the concept of limits in two dimensions, particularly around critical points
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Students and educators in mathematics, particularly those studying calculus and analysis, as well as researchers exploring properties of multivariable functions and their applications.

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The question looks like this.
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4 $.

(a) Show that $f$ is discontinuous at (0, 0)

(b) Show that $f$ is discontinuous on two entire curves.
In regarding (a), I know $f(x, y)$ is discontinuous on certain directions, but can't elaborate it in decent form.

In regarding (b), How can I show it?
 
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V150 said:
Let $f(x, y)$ = 0 if $y\leq 0$ or $y\geq x^4$, and $f(x, y)$ = 1 if $0 < y < x^4 $.

(a) Show that $f$ is discontinuous at (0, 0)
By definition, $f$ is continuous at $(0,0)$ if for every $\varepsilon>0$ there exists a $\delta>0$ such that if $(x,y)$ is located within distance $\delta$ from $(0,0)$, then $|f(x,y)-f(0,0)|=|f(x,y)|<\varepsilon$. Choose any $0<\varepsilon<1$. Can you find a circle around $(0,0)$ such that within that circle $f$ takes values $<\varepsilon$?

V150 said:
(b) Show that $f$ is discontinuous on two entire curves.
By the two curves, does the problem mean $y(x)=0$ and $y(x)=x^4$? Again, within every circle whose center lies on these curves, however small the radius is, there are points where $f$ returns 0 and other points where $f$ returns 1.
 

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