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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?

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    $\begingroup$ No there isn't. $\endgroup$ – Augustin Aug 4 '15 at 9:02
  • $\begingroup$ Discontinuit fot the function at the origin? Your book is wrong! $\endgroup$ – Euler88 ... Aug 4 '15 at 9:02
  • $\begingroup$ The book is wrong $\endgroup$ – David Quinn Aug 4 '15 at 9:02
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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 $x=0$.

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  • $\begingroup$ Very-very long and sound for such trivial problem. $\endgroup$ – Michael Galuza Aug 4 '15 at 9:18
  • $\begingroup$ But for proving continuity or discontinuity at $x=a$, we have to prove $LHL=RHL=f(a)$ . Isn't it right? $\endgroup$ – Harish Chandra Rajpoot Aug 4 '15 at 9:24
  • $\begingroup$ Great thank you! So it must have been wrong in the textbook. $\endgroup$ – Bob Aug 4 '15 at 9:26
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    $\begingroup$ @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. $\endgroup$ – David K Aug 6 '15 at 13:29
  • $\begingroup$ @DavidK, textbook is not Pope $\endgroup$ – Michael Galuza Aug 6 '15 at 13:30
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There is no discontinuity at the origin.

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