What is the limit of a rational function as it approaches its vertical asymptote?

Question

For example, take the function $f(x)=\frac{1}{(x-3)^2}$. What is the the limit as x approaches 3? (sorry, I don't know how to format this question)

My teacher says that there is no limit at x=3, since infinity is not a discrete value, but one of my classmate argues that otherwise, although I forgot their exact reasoning. Can someone please explain who is correct? ${}{}{}{}{}{}{}{}{}$

Answer 1

It is true that the limit does not exist at this vertical asymptote, but we can be a bit more specific about it. In the case of the function you gave, the limit as $x$ approaches three is $\infty$, since the function grows without bound when coming from the left and the right. Note that since infinity is not a number, the limit still does not exist. It is just that in certain cases when the left and right sides approach the same value or similar direction can we be more specific.

Answer 2

If you agree with the fact that, as $u \to 0$, you get $$ \frac 1{u^2} \to +\infty, $$ setting $u:=x-3$, then as $x \to 3$ we have $u \to 0$ and $$ f(x)=\frac{1}{(x-3)^2}=\frac 1{u^2} \longrightarrow +\infty $$ and your teacher has the correct answer.