2.3 The limit laws

1. The limit of x as x approaches a is a : $\underset{x\to 2}{\text{lim}}x=2.$
2. The limit of a constant is that constant: $\underset{x\to 2}{\text{lim}}5=5.$

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

$\begin{array}{ccccc}\underset{x\to \mathrm{-3}}{\text{lim}}\left(4x+2\right)\hfill & =\underset{x\to \mathrm{-3}}{\text{lim}}4x+\underset{x\to \mathrm{-3}}{\text{lim}}2\hfill & & & \text{Apply the sum law.}\hfill \\ & =4·\underset{x\to \mathrm{-3}}{\text{lim}}x+\underset{x\to \mathrm{-3}}{\text{lim}}2\hfill & & & \text{Apply the constant multiple law.}\hfill \\ & =4·\left(\mathrm{-3}\right)+2=\mathrm{-10}.\hfill & & & \text{Apply the basic limit results and simplify.}\hfill \end{array}$

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

$\begin{array}{}\\ \\ \underset{x\to 2}{\text{lim}}\frac{2x^2-3x+1}{x^3+4}\hfill & =\frac{\underset{x\to 2}{\text{lim}}\left(2x^2-3x+1\right)}{\underset{x\to 2}{\text{lim}}\left(x^3+4\right)}\hfill & & & \text{Apply the quotient law, making sure that.}{\left(2\right)}^3+4\ne 0\hfill \\ & =\frac{2·\underset{x\to 2}{\text{lim}}{x}^2-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{\underset{x\to 2}{\text{lim}}{x}^3+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the sum law and constant multiple law.}\hfill \\ & =\frac{2·{\left(\underset{x\to 2}{\text{lim}}x\right)}^2-3·\underset{x\to 2}{\text{lim}}x+\underset{x\to 2}{\text{lim}}1}{{\left(\underset{x\to 2}{\text{lim}}x\right)}^3+\underset{x\to 2}{\text{lim}}4}\hfill & & & \text{Apply the power law.}\hfill \\ & =\frac{2\left(4\right)-3\left(2\right)+1}{{\left(2\right)}^3+4}=\frac{1}{4}.\hfill & & & \text{Apply the basic limit laws and simplify.}\hfill \end{array}$