Find the following limits, rationalizing the numerator where necessary.

Example: $\lim\limits _ { x \rightarrow 1 } \frac { \sqrt { x } - 1 } { x - 1 } = \lim\limits _ { x \rightarrow 1 } \frac { \sqrt { x } - 1 } { x - 1 } \cdot \frac { \sqrt { x } + 1 } { \sqrt { x + 1 } } = \lim\limits _ { x \rightarrow 1 } \frac { x - 1 } { ( x - 1 ) \sqrt { x + 1 } } = \lim\limits _ { x \rightarrow 1 } \frac { 1 } { \sqrt { x + 1 } } = \frac { 1 } { 2 }$

1. $\lim\limits _ { x \rightarrow 4 } \frac { \sqrt { x } - 2 } { x - 4 }$

   Hint (a):

   After rationalizing the numerator, did you get:

   $\lim\limits_{ x \to 4 }\frac{x-4}{(x-4)(\sqrt{x}+2)}$

2. $\lim\limits _ { x \rightarrow 6 } \frac { \sqrt { x + 2 } - \sqrt { 2 } } { x }$

   Hint (b):

   Is it necessary to rationalize the numerator in this case?
   What happens when you just use substitution?

3. $\lim\limits_ { x \rightarrow 2 } \frac { x ^ { 3 } - 8 } { x - 2 }$

   Hint (c):

   Recall the cubic factoring formula:

   $\textit{a}^3-\textit{b}^3=(\textit{a}-\textit{b})(\textit{a}^2+\textit{ab}+\textit{b}^2)$