Question

evaluate the limit, if it exists.
a) (limlimits_{t \to 2^-} left( \frac{3t^3 - 2t^2 - 8t}{3t^2 - 12t + 12}
ight))

Explanation:

Step1: Factor numerator and denominator

First, factor the numerator \(3t^{3}-2t^{2}-8t\) and the denominator \(3t^{2}-12t + 12\).

For the numerator:
\(3t^{3}-2t^{2}-8t=t(3t^{2}-2t - 8)\)
Factor \(3t^{2}-2t - 8\): we need two numbers that multiply to \(3\times(-8)=-24\) and add to \(-2\). The numbers are \(-6\) and \(4\).
So \(3t^{2}-2t - 8=3t^{2}-6t + 4t - 8 = 3t(t - 2)+4(t - 2)=(3t + 4)(t - 2)\)
Thus, the numerator is \(t(3t + 4)(t - 2)\)

For the denominator:
\(3t^{2}-12t + 12=3(t^{2}-4t + 4)=3(t - 2)^{2}\) (since \(t^{2}-4t + 4=(t - 2)^{2}\))

So the function becomes \(\frac{t(3t + 4)(t - 2)}{3(t - 2)^{2}}=\frac{t(3t + 4)}{3(t - 2)}\) (for \(t
eq2\))

Step2: Analyze the limit as \(t

ightarrow2^{-}\)
We now find the limit as \(t\) approaches \(2\) from the left (\(t
ightarrow2^{-}\)). Let's substitute values close to \(2\) from the left (e.g., \(t = 1.9\), \(t = 1.99\)) into \(\frac{t(3t + 4)}{3(t - 2)}\)

First, consider the sign of the denominator: when \(t
ightarrow2^{-}\), \(t - 2<0\) (since \(t\) is less than \(2\))

The numerator: when \(t
ightarrow2^{-}\), \(t>0\) (close to \(2\)) and \(3t + 4>0\) (since \(t>0\)), so the numerator \(t(3t + 4)>0\)

So the fraction \(\frac{\text{positive}}{\text{negative}\times3}\) (since denominator is \(3(t - 2)\) and \(t - 2<0\)) will be negative.

Now, let's find the limit by substituting \(t = 2\) (after canceling the non - zero factor \((t - 2)\)) into the simplified function, but considering the sign.

Substitute \(t = 2\) into \(\frac{t(3t + 4)}{3(t - 2)}\): numerator is \(2\times(3\times2 + 4)=2\times(6 + 4)=2\times10 = 20\), denominator is \(3\times(2 - 2)=0\). But we are approaching from the left, so we analyze the behavior.

As \(t
ightarrow2^{-}\), \(t - 2
ightarrow0^{-}\), so \(\frac{t(3t + 4)}{3(t - 2)}
ightarrow\frac{20}{3\times0^{-}}\). Since we have a non - zero numerator and the denominator approaches \(0\) from the negative side, the limit will be \(-\infty\)

Wait, let's re - check the factoring:

Wait, numerator: \(3t^{3}-2t^{2}-8t=t(3t^{2}-2t - 8)\). Let's factor \(3t^{2}-2t - 8\) again. Using the quadratic formula for \(ax^{2}+bx + c = 0\), \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(3t^{2}-2t - 8\), \(a = 3\), \(b=-2\), \(c = - 8\). Then \(t=\frac{2\pm\sqrt{4+96}}{6}=\frac{2\pm\sqrt{100}}{6}=\frac{2\pm10}{6}\). So \(t=\frac{2 + 10}{6}=2\) or \(t=\frac{2-10}{6}=-\frac{4}{3}\). So \(3t^{2}-2t - 8=3(t - 2)(t+\frac{4}{3})=(3t + 4)(t - 2)\) (since \(3(t+\frac{4}{3})=3t + 4\)). So numerator is \(t(3t + 4)(t - 2)\), denominator is \(3(t - 2)^{2}\). So the simplified function is \(\frac{t(3t + 4)}{3(t - 2)}\) for \(t
eq2\).

Now, as \(t
ightarrow2^{-}\), let's take \(t = 2 - h\) where \(h
ightarrow0^{+}\) (since \(t
ightarrow2^{-}\), \(h\) is a small positive number).

Substitute \(t = 2 - h\) into \(\frac{t(3t + 4)}{3(t - 2)}\):

\(t=2 - h\), \(3t + 4=3(2 - h)+4=6-3h + 4=10 - 3h\), \(t - 2=-h\)

So the function becomes \(\frac{(2 - h)(10 - 3h)}{3(-h)}=\frac{(2 - h)(10 - 3h)}{-3h}\)

As \(h
ightarrow0^{+}\), the numerator \((2 - h)(10 - 3h)
ightarrow2\times10 = 20\) (a positive number), and the denominator \(-3h
ightarrow0^{-}\) (a negative number approaching \(0\)). So the fraction \(\frac{20}{-3h}\) (where \(h
ightarrow0^{+}\)) will approach \(-\infty\) because we have a positive numerator and a negative denominator that is getting closer to \(0\), so the magnitude of the fraction becomes very large in the negative direction.

Answer:

The limit as \(t
ightarrow2^{-}\) of \(\frac{3t^{3}-2t^{2}-8t}{3t^{2}-12t + 12}\) is \(-\infty\)