How do you find the limit of `|x+2| / (x+2)` as x approaches `-2`?

The absolute value function `abs(x+2)` can be defined as the piecewise function

`abs(x+2)={(x+2,;,x>=-2),(-(x+2),;,x<-2):}`

We should determine if the limit from the left approaches the limit from the right.

Limit from the left:

When the function is directly to the left of `x=-2`, we are on the `-(x+2)` portion of the piecewise function since `x<-2`.

Thus, the function when `x<-2` becomes

`abs(x+2)/(x+2)=(-(x+2))/(x+2)=-1`

Hence the limit from the left is

`lim_(xrarr-2^-)abs(x+2)/(x+2)=lim_(xrarr-2^-)-1=-1`

Limit from the right:

From the right, `x> -2`, so we just use `x+2` in place of `abs(x+2)`.

The function becomes

`abs(x+2)/(x+2)=(x+2)/(x+2)=1`

So the limit from the right is

`lim_(xrarr-2^+)abs(x+2)/(x+2)=lim_(xrarr-2^+)1=1`

Relating the limits:

Since `lim_(xrarr-2^+)abs(x+2)/(x+2)!=lim_(xrarr-2^-)abs(x+2)/(x+2)`, we know that `lim_(xrarr-2)abs(x+2)/(x+2)` does not exist.

In fact, we saw that the function is simply `y=-1` when `x<-2` and is `y=1` when `x> -2`.

Graphed is `abs(x+2)/(x+2)`:

graph{abs(x+2)/(x+2) [-13.77, 8.73, -5.62, 5.63]}