Option 9: Vertical Asymptotes (Limits that equal infinity)

Definition: Vertical Asymptotes

Vertical asymptotes occur at an $x–\mathit{value}$ when the answer to a one-sided limit (left-hand limit or right-hand limit) at that $x–\mathit{value}$  is $\mathit{\pm }\mathit{\infty }$.

LHL : $\underset{x\to a^{–}}{\lim }f\left(\mathrm{(}x\mathrm{)}\right)=\mathbf{+}\mathbf{\infty }$ or $\underset{x\to a^{–}}{\lim }f\left(\mathrm{(}x\mathrm{)}\right)=\mathit{–}\mathit{\infty }$

RHL : $\underset{x\to a^{+}}{\lim }f\left(\mathrm{(}x\mathrm{)}\right)=\mathbf{+}\mathbf{\infty }$ or $\underset{x\to a^{+}}{\lim }f\left(\mathrm{(}x\mathrm{)}\right)=\mathit{–}\mathit{\infty }$

Note:

–           The answer to a vertical asymptote question is always the $x–\mathit{value}$ where it is occurring.

–           The overall limit does not need to exist in order for there to be a vertical asymptote. The left-hand limit does not need to equal the right-hand limit.

–           Only one of the limits (left or right) needs to equal $\mathit{\pm }\mathit{\infty }$. They both do not need to equal $\mathit{\pm }\mathit{\infty }$.

–           Vertical asymptotes most often occur where you would get division by zero in your equation. They do not always occur there. Remember that if you are able to cancel out a part of an equation, then you have a removeable discontinuity (hole in the graph) not a vertical asymptote.

For Example : In this equation you would have a vertical asymptote at $x=–9$ because can’t get rid of that part of the equation. However, you would have a removeable discontinuity at $x=8$ because the $\left(\mathrm{(}x–8\mathrm{)}\right)$ terms can be canceled.

$\underset{x\to 8}{\lim }\frac{\overline{)\left(\mathrm{(}x–8\mathrm{)}\right)}\left(\mathrm{(}x+2\mathrm{)}\right)}{\overline{)\left(\mathrm{(}x–8\mathrm{)}\right)}\left(\mathrm{(}x+9\mathrm{)}\right)}$