Defining Derivatives with Limits Flashcards - GRE Subject Test: Math

Step 1: Try plugging in into the denominator of the function. We want to make sure that the bottom does not become ...

.. We got zero, and we cannot have zero in the denominator. So, we must try and factor the function (numerator and denominator):

Step 2: Factor:

$\frac {(x+2)(x-2)}{(x-2)(x^2+2x+4)}$

Step 3: Reduce:

$\frac {x+2}{x^2+2x+4}$

Step 4: Now that we got rid of the factor that made the denominator zero, we know that this function has a limit.

Step 5: Plug in into the reduced factor form:

Simplify as much as possible...

$\frac {2+2}{2^2+2(2)+4}=\frac {4}{12}=\frac {1}{3}$

The limit of this function as x approaches is $\frac {1}{3}$

Defining Derivatives with Limits - GRE Subject Test: Math