Question

Find the given limits if they exist. If a limit does not exist, explain why.

$\underset{t\to 0}{\lim }\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(3+t\right)-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$}\right.}}{t},\frac{{\left(3+t\right)}^2-9}{t}\right)$.

Short answer

The limit exists. The solution is,

$\left(-\frac{1}{3},6\right)$.

Step-by-step solution

Step 1. Given Information.

The limit is,

$\underset{t\to 0}{\lim }\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(3+t\right)-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$}\right.}}{t},\frac{{\left(3+t\right)}^2-9}{t}\right)$.

Step 2. Whether the limit exists or not.

Consider the limit,

$\underset{t\to 0}{\lim }\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(3+t\right)-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$}\right.}}{t},\frac{{\left(3+t\right)}^2-9}{t}\right)$

The objective is to evaluate the above limit.

$$\begin{gathered} \underset{t\to t_0}{\lim }r\left(t\right)=\underset{t\to t_0}{\lim }\left(x\right(t),y(t),z(t\left)\right) \\ =\left(\underset{t\to t_0}{\lim }x\right(t),\underset{t\to t_0}{\lim }y(t),\underset{t\to t_0}{\lim }z(t\left)\right) \end{gathered}$$

Here, $\underset{t\to t_0}{\lim }x\left(t\right),\underset{t\to t_0}{\lim }y\left(t\right),$and $\underset{t\to t_0}{\lim }z\left(t\right)$exist.

Step 3. Finding the limit.

$$\begin{gathered} \underset{t\to 0}{\lim }\left(\frac{1}{t}\left(\frac{1}{\left(3+t\right)}-\frac{1}{3}\right),\frac{{\left(3+t\right)}^2-9}{t}\right)=\underset{t\to 0}{\lim }\left(\frac{1}{t}\left(\frac{3-3-t}{(3+t)}\right),\frac{9+t^2+6t-9}{t}\right) \\ =\underset{t\to 0}{\lim }\left(\frac{1}{t}\left(\frac{-t}{3+t}\right),\frac{t^2+6t}{t}\right) \\ =\underset{t\to 0}{\lim }\left(\left(\frac{-1}{3+t}\right),t+6\right) \\ =\left(\underset{t\to 0}{\lim }\left(\frac{-1}{3+t}\right),\underset{t\to 0}{\lim }\left(t+6\right)\right) \\ =\left(-\frac{1}{3},6\right) \end{gathered}$$