Example 4: Vertical Asymptotes

$\underset{x\to 2}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}$

Step 1: Decide if you are going to run a left-hand limit right-hand limit, or overall limit.

This example is specifically asking for an overall limit.

$\underset{x\to 2}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}$

Step 2: Try Option 1: Plug it in

Always try plugging in the $x–\mathit{value}$ you are heading towards.

In this example we get division by zero, which means Option 1 has failed.

$\underset{x\to 2}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}\left(2\right)–2\mathrm{)}\right)}^2}=\frac{1}{0}$

Step 3: Try Option 2: Factor and Cancel

No factoring or canceling is possible in this example.

$\underset{x\to 2}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}$

Step 4: Back to Option 1: Plug It In with an $x–\mathit{value}$ that is really, really, really close to the $x–\mathit{value}$ that you actually care about.

You will need to use that $x–\mathit{value}$ to approximate the value on the top of your fraction and the bottom of your fraction.

In this overall limit example, we would need to choose a number to the left and to the right of $x=2$that is really, really, really close to the $x=2$. In this example I am going to use $x=1.999999999$ and $x=2.000000001$.

Remember the actual $x–\mathit{value}$ doesn’t really matter because you aren’t really doing this on a piece of paper, it is all a conceptual process in your head.

The major difference between this problem and the previous examples is that we are squaring the bottom term. This squaring causes both the left– and right-hand limit to be positive.

LHL:
$\underset{x\to 2^{–}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}(1.999999999)–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}\mathbf{–}.000000001\mathrm{)}\right)}^2}=\mathit{+}1,000,000,000,000,000,000$ 

$\underset{x\to 2^{–}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}(1.999999999)–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}\mathbf{–}.000000001\mathrm{)}\right)}^2}=\mathit{+}\mathit{Huge\; Number}$

RHL:

$\underset{x\to 2^{+}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}(2.000000001)–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}\mathit{+}.000000001\mathrm{)}\right)}^2}=\mathit{+}1,000,000,000,000,000,000$

$\underset{x\to 2^{+}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}(2.000000001)–2\mathrm{)}\right)}^2}=\frac{1}{{\left(\mathrm{(}\mathit{+}.000000001\mathrm{)}\right)}^2}=+\mathit{Huge\; Number}$

Step 5: Use logic to evaluate the final result from Step 4.

LHL:
$\underset{x\to 2^{–}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{+\mathit{Big}}{+\mathit{Small}}=+\mathit{Huge\; Number}$ 

$\underset{x\to 2^{–}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{+\mathit{Big}}{+\mathit{Small}}=+\infty$

RHL:
$\underset{x\to 2^{+}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{+\mathit{Big}}{+\mathit{Small}}=+\mathit{Huge\; Number}$  

$\underset{x\to 2^{+}}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=\frac{+\mathit{Big}}{+\mathit{Small}}=+\infty$

$\mathbf{LHL}\mathbf{=}\mathbf{RHL}$

$+\infty =+\infty$

Final Result:

$\underset{x\to 2}{\lim }\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}=+\infty$

Meaning:

The overall limit as $x$ approaches $2$ of $\frac{1}{{\left(\mathrm{(}x–2\mathrm{)}\right)}^2}$ is $y=+\infty$.

There is a vertical asymptote at $x=2$.

Remember you only need a one-sided limit to equal $\mathit{\pm }\mathit{\infty }$for there to be a vertical asymptote.