Example 2: Vertical Asymptotes

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

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 a right-hand limit.

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

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.

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

Step 3: Try Option 2: Factor and Cancel

No factoring or canceling is possible in this example.

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

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 example we need a number to the right of $x=2$that is really, really, really close to the $x=2$. In this example I am going to use $x=2.000000001$.

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 top of the fraction would be considered a $+\mathit{Big}$ since 1 is a large number compared to a decimal.

The bottom of the fraction would be considered a $+\mathit{Small}$. Since your $x–\mathit{value}$that you use will always be larger than the 2, the result will always be a positive decimal.

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

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

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

Here we have a $+\mathit{Big}$divided by a$+\mathit{Small}$, which means the number is getting huge, and since we have $\frac{\mathbf{+}}{+}=+\mathbf{Huge\; Number}$.

So, $+\infty$.

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

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

Final Result:

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

Meaning:

The right-hand limit as $x$  approaches $2$ of $\frac{1}{\left(\mathrm{(}x–2\mathrm{)}\right)}$ 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.