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For the function $$y = \frac{2x^2}{x-3},$$ I understand that $x = 3$ and $y = 2x+6$ are asymptote, but according to the answers in my textbook, there is a discontinuity at the origin. Why is this?

Harish Chandra Rajpoot
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Bob
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2 Answers2

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There is no discontinuity at the origin.

davidlowryduda
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For, discontinuity at $x=0$, let's check both left hand & right hand limits as follows

$$LHL=\lim_{x\to 0^{-}}\frac{2x^2}{x-3}$$ $$=\lim_{h\to 0}\frac{2(0-h)^2}{(0-h)-3}$$ $$=\lim_{h\to 0}\frac{2h^2}{-h-3}$$ $$=\lim_{h\to 0}\frac{-2h^2}{h+3}=\frac{-2(0)^2}{0+3}=0$$

Similarly, $$RHL=\lim_{x\to 0^{+}}\frac{2x^2}{x-3}$$ $$=\lim_{h\to 0}\frac{2(0+h)^2}{(0+h)-3}$$ $$=\lim_{h\to 0}\frac{2h^2}{h-3}$$ $$=\frac{2(0)^2}{0+3}=0$$ & we have $$f(0)=\frac{2(0)^2}{0-3}=0$$ Thus, we have $$LHL=RHL=f(0)$$ Hence the function $y=\frac{2x^2}{x-3}$ is continuous at $x=0$ i.e. it has no discontinuity at $x=0$.

Harish Chandra Rajpoot
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  • Very-very long and sound for such trivial problem. – Michael Galuza Aug 04 '15 at 09:18
  • But for proving continuity or discontinuity at $x=a$, we have to prove $LHL=RHL=f(a)$ . Isn't it right? – Harish Chandra Rajpoot Aug 04 '15 at 09:24
  • Great thank you! So it must have been wrong in the textbook. – Bob Aug 04 '15 at 09:26
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    @MichaelGaluza Yes, this is rather like swatting a fly with a sledgehammer. But when the "fly" is a statement made with the authority of a textbook, it becomes a big, big, hairy fly, and to make sure you really kill it, the sledgehammer may be a good tool. – David K Aug 06 '15 at 13:29
  • @DavidK, textbook is not Pope – Michael Galuza Aug 06 '15 at 13:30
  • @MichaelGaluza Of course no textbook is the Pope, but from the point of view of a student, when the student's understanding contradicts the book, the error is almost always on the student's side. At first glance, then, "the book is wrong" is an extraordinary claim for the student to believe, and so extraordinary evidence is merited. – David K Aug 06 '15 at 13:41