Example 1: Absolute Value Limit

$\underset{y\to –1}{\lim }\frac{y^2–1}{\left||2y+2|\right|}$

Step 1: 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.

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

Step 2: Split the equation into its two equations. A left-hand limit and a right-hand limit.

$\underset{y\to –1}{\lim }\frac{y^2–1}{\left||2y+2|\right|}$

LHL: Drop the absolute value bars and replace them with a $–\left(\mathrm{ }\right).$

$\underset{y\to {–1}^{–}}{\lim }\frac{y^2–1}{–(2y+2)}=$

RHL: Drop the absolute value bars

$\underset{y\to {–1}^{+}}{\lim }\frac{y^2–1}{2y+2}=$

Step 3: Determine the overall limit by finding your left-hand limit and right-hand limit using the equations you created in Step 2.

To solve both the LHL and the RHL we will use two previous limit options. First, we will use Option 2: Factor and Cancel, and then back to Option 1: Plug it in.

$\underset{y\to {–1}^{–}}{\lim }\frac{y^2–1}{–(2y+2)}=$

$\underset{y\to {–1}^{–}}{\lim }\frac{(y–1)(y+1)}{–(2y+2)}=$

$\underset{y\to {–1}^{–}}{\lim }\frac{(y–1)(y+1)}{–2(y+1)}=$

$\underset{y\to {–1}^{–}}{\lim }\frac{(y–1)\overline{)(y+1)}}{–2\overline{)(y+1)}}=$

$\underset{y\to {–1}^{–}}{\lim }\frac{(y–1)}{–2}=$

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

$\underset{y\to {–1}^{+}}{\lim }\frac{y^2–1}{2y+2}=$

$\underset{y\to {–1}^{+}}{\lim }\frac{(y–1)(y+1)}{2y+2}=$

$\underset{y\to {–1}^{+}}{\lim }\frac{(y–1)(y+1)}{2(y+1)}=$

$\underset{y\to {–1}^{+}}{\lim }\frac{(y–1)\overline{)(y+1)}}{2\overline{)(y+1)}}=$

$\underset{y\to {–1}^{+}}{\lim }\frac{(y–1)}{2}=$

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

$\mathbf{LHL}\ne \mathbf{RHL}$

$\mathit{1}\ne \mathbf{–}\mathbf{1}$

$\underset{y\to –1}{\lim }\frac{y^2–1}{\left||2y+2|\right|}=\mathit{Does\; Not\; Exist}$

Final Result:

$\underset{y\to –1}{\lim }\frac{y^2–1}{\left||2y+2|\right|}=\mathit{Does\; Not\; Exist}$

Meaning:

The overall limit as $y$ approaches $–1$ of $\frac{y^2–1}{\left||2y+2|\right|}$ Does Not Exist.