evaluate the limit using the appropriate limit law(s). (if an answer does not exist, enter…

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

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

Answer

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\rightarrow a}f(t)=L_1) and (\lim_{t\rightarrow a}g(t)=L_2\neq0), then (\lim_{t\rightarrow 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\rightarrow - 2}\frac{t^{4}-8}{2t^{2}-3t + 1}=\frac{8}{15}).

Answer:

(\frac{8}{15})