Does the graph of the function $\frac{x(1+x)}{2x^2}$ have a hole or a vertical asymptote at $x=0$?

Question

From what I remember, for the function $\frac{x(1+x)}{2x^2}$, when we cross out a factor in the numerator and denominator, we get a hole. So, then we get $\frac{1+x}{2x}$. But then it seems that $x=0$ is the vertical asymptote when I thought $x=0$ was supposed to be a hole. I was hoping that someone could clarify my confusion.

Answer 1

You have an asymptote at $x=0$ when one or both of the one-sided limits of the function, as $x$ approaches $0$, is $\infty$ or $-\infty$.

For the function in question,

$$ \lim_{x\to 0^+}\frac{x(1+x)}{2x^2} = \lim_{x\to 0^+}\frac{1+x}{2x} = \lim_{x\to 0^+}\frac{1}{2x}+\frac{1}{2} = \infty+\frac{1}{2} = \infty $$ so in this case, it's an asymptote, not a hole.

If instead, the function was $$\frac{x(1+x)}{2x}$$ then at $x=0$, we would get $$\lim_{x\to 0}\frac{x(1+x)}{2x} = \lim_{x\to 0}\frac{1+x}{2} = \lim_{x\to 0}\frac{1}{2}+\frac{x}{2} = \frac{1}{2}+0 = \frac{1}{2} $$ so the limit at $x=0$ exists (and is finite), but since the function is not defined at $x=0$, the graph has a "hole" at the point $(0,\frac{1}{2})$.

Answer 2

As you can see, we do have a vertical asymptote, not a hole. As one of the comments said, if a function is "blowing" up while approaching 0, it is an asymptote. Now, the other thing you did wrong was crossing out the factor. What you actually did is 0/0, getting you to a wrong assumption.

Answer 3

The graph of $$\frac{x(1+x)}{2x^2}$$seems to have a hole at $x=0.$

After cancelling the x from top and bottom we get $$\frac{(1+x)}{2x}$$ which has a vertical asymptote.

Thus it has a vertical asymptote at $x=0.$

Answer 4

One way to see where the asymptote comes from is to note that $$ \frac{x(1+x)}{2x^2} = \frac{x + x^2}{2x^2} = \frac{1}{2x} + \frac{1}{2}, $$ which is the standard hyperbola $y=1/x$, scaled down by $2$ (which does not affect the asymptotes here) and translated up $1/2$, which moves the horizontal asymptotes up $1/2$ but has no effect on vertical asymptotes.