Question

How do you find the limit of `\lim _ { x \rightarrow - 2} \frac { x + 2} { x ^ { 2} + 6x + 8}`?

Answer 1

`lim_(xrarr-2)(x+2)/(x^2+6x+8)=1/2`

Explanation:

Notice that plugging in `x=-2` in the function's current form gives `(-2+2)/((-2)^2+6(-2)+8)=0/0`, so the limit is currently in an indeterminate form and we cant yet determine the limit.

However, notice that we can factor the denominator of the fraction:

`lim_(xrarr-2)(x+2)/(x^2+6x+8)=lim_(xrarr-2)(x+2)/((x+2)(x+4))`

Now we can see why plugging in `x=-2` into the denominator creates a `0`. However, we can cancel the `(x+2)` terms in the numerator and denominator:

`=lim_(xrarr-2)1/(x+4)`

In other words, the function `(x+2)/(x^2+6x+8)` is identical to the function `1/(x+4)` except at the point `x=-2`. The former function is identical to the latter function except that the first function has a hole at `x=-2`. So, the limit is where the function should lie, which can be found by plugging `x=-2` into the simplified function, `1/(x+4)`.

`=1/(-2+4)=1/2`