Question

evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter dne.)
\\(\lim_{t\to - 2}\frac{t^{4}-8}{2t^{2}-3t + 1}\\)

Explanation:

Step1: Substitute \(t = - 2\) into the function

First, find the value of the numerator \(t^{4}-8\) and the denominator \(2t^{2}-3t + 1\) when \(t=-2\).
For the numerator: \(t^{4}-8=(-2)^{4}-8=16 - 8=8\).
For the denominator: \(2t^{2}-3t + 1=2\times(-2)^{2}-3\times(-2)+1=2\times4 + 6+1=8 + 6+1=15\).

Step2: Use the quotient - limit law

The quotient - limit law states that if \(\lim_{t
ightarrow a}f(t)=L_1\) and \(\lim_{t
ightarrow a}g(t)=L_2
eq0\), then \(\lim_{t
ightarrow a}\frac{f(t)}{g(t)}=\frac{L_1}{L_2}\). Here, \(f(t)=t^{4}-8\), \(g(t)=2t^{2}-3t + 1\), \(a=-2\), \(L_1 = 8\) and \(L_2=15\).
So \(\lim_{t
ightarrow - 2}\frac{t^{4}-8}{2t^{2}-3t + 1}=\frac{8}{15}\).

Answer:

\(\frac{8}{15}\)