To determine the limit as [tex]\( t \)[/tex] approaches 2 for the given piecewise function, we need to examine the function's behavior from different directions (left and right of [tex]\( t = 2 \)[/tex]) as well as directly at [tex]\( t = 2 \)[/tex].
The piecewise function is defined as:
[tex]\[ f(t) = \begin{cases}
t^2 & \text{if } t < -2 \\
\frac{t + 6}{t^2 - t} & \text{if } -1 < t < 2 \\
3t - 2 & \text{if } t \geq 2
\end{cases} \][/tex]
1. Approaching [tex]\( t = 2 \)[/tex] from the left ([tex]\( t \to 2^- \)[/tex]):
For [tex]\( -1 < t < 2 \)[/tex], the function is defined as [tex]\( f(t) = \frac{t+6}{t^2 - t} \)[/tex]. We need to find the limit
[tex]\[
\lim_{t \to 2^-} \frac{t+6}{t^2 - t}.
\][/tex]
Since [tex]\( \frac{t+6}{t^2 - t} \)[/tex] is continuous and defined for [tex]\( -1 < t < 2 \)[/tex], we can directly substitute [tex]\( t = 2 \)[/tex]:
[tex]\[
\frac{2+6}{2^2 - 2} = \frac{8}{4 - 2} = \frac{8}{2} = 4.
\][/tex]
2. Directly at [tex]\( t = 2 \)[/tex] (within the interval [tex]\( -1 < t < 2 \)[/tex]):
From the previous step, the same expression [tex]\( \frac{t+6}{t^2 - t} \)[/tex] applies:
[tex]\[
\frac{2 + 6}{2^2 - 2} = \frac{8}{4 - 2} = \frac{8}{2} = 4.
\][/tex]
Thus, the limit as [tex]\( t \)[/tex] directly approaches 2 is also [tex]\( 4 \)[/tex].
3. Approaching [tex]\( t = 2 \)[/tex] from the right ([tex]\( t \to 2^+ \)[/tex]):
For [tex]\( t \geq 2 \)[/tex], the function is defined as [tex]\( f(t) = 3t - 2 \)[/tex]. We need to find the limit
[tex]\[
\lim_{t \to 2^+} (3t - 2).
\][/tex]
Since [tex]\( 3t - 2 \)[/tex] is continuous and defined for [tex]\( t \geq 2 \)[/tex], we can directly substitute [tex]\( t = 2 \)[/tex]:
[tex]\[
3 \cdot 2 - 2 = 6 - 2 = 4.
\][/tex]
In conclusion, the limit of the piecewise function as [tex]\( t \)[/tex] approaches 2 from any direction is:
[tex]\[
\boxed{4}
\][/tex]
