Limit Laws

$\lim\limits_{x \to a}\ \left( f(x) + g(x) \right) = \lim\limits_{x \to a}\ f(x) + \lim\limits_{x \to a}\ g(x).$

$$\lim\limits_{x \to a}\ \left(f(x) \ g(x) \right) = \left(\lim\limits_{x \to a}\ f(x)\right) \left(\lim\limits_{x \to a}\ g(x)\right).$$

$$\lim\limits_{x \to a}\ \dfrac{f(x) }{ g(x) } = \dfrac{\lim\limits_{x \to a}\ f(x) }{\lim\limits_{x \to a}\ g(x) }.$$

$$\lim\limits_{x \to a}\ f \left( g(x) \right) = f \left( \lim\limits_{x \to a}g(x) \right),$$ assuming $f$ is a continuous function.

$$\lim\limits_{x \to a}\ 17 = 17.$$

$$\lim\limits_{x \to a}\ \left( f(x) \right)^2 = \left( \lim\limits_{x \to a}f(x) \right)^2.$$

$$\lim\limits_{x \to a}\ \left( f(x) \right)^n = \left( \lim\limits_{x \to a} f(x) \right)^n .$$

\begin{eqnarray} \lim\limits_{x \to a}\ (7x - 2)^3 &=& \left(\lim\limits_{x \to a}\ (7x - 2) \right)^3 \\ \ &=& (7a - 2)^3 . \end{eqnarray}

Since the square root function is continuous, \begin{eqnarray} \lim\limits_{x \to 0}\ \sqrt{x+4} &=& \sqrt{\lim\limits_{x \to 0}\ (x+4)} \\ \ &=& \sqrt {4} \\ \ &=& 2. \end{eqnarray}

\begin{eqnarray} \lim\limits_{x \to\ -7}\ \sqrt{x+4} &=& \sqrt{-7 + 4} \\ \ &=& \sqrt{-3} \\ \ &=& i \sqrt{3}, \end{eqnarray} where $i = \sqrt{-1}$ is called the "imaginary number."