Why is there a discontinuity at zero?

Question

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?

Answer 1

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 $x=0$.

Answer 2

There is no discontinuity at the origin.