use lhôpitals rule to find the limit. lim t→ - 2 (t³ - 3t + 2)/(t² - t - 6) lim t→ - 2 (t³ - 3t + 2)/(t² - t…

Explanation:

Step1: Check indeterminate form

Substitute (t = - 2) into (\frac{t^{3}-3t + 2}{t^{2}-t - 6}). (\frac{(-2)^{3}-3\times(-2)+2}{(-2)^{2}-(-2)-6}=\frac{-8 + 6+2}{4 + 2-6}=\frac{0}{0}), which is an indeterminate - form. So, we can apply L'Hopital's Rule.

Step2: Differentiate numerator and denominator

Differentiate the numerator (y_1=t^{3}-3t + 2) using the power - rule ((x^n)^\prime=nx^{n - 1}). (y_1^\prime=3t^{2}-3). Differentiate the denominator (y_2=t^{2}-t - 6), (y_2^\prime=2t-1).

Step3: Find the new limit

Now, find (\lim_{t\rightarrow - 2}\frac{3t^{2}-3}{2t - 1}). Substitute (t=-2) into (\frac{3t^{2}-3}{2t - 1}), we get (\frac{3\times(-2)^{2}-3}{2\times(-2)-1}=\frac{3\times4-3}{-4 - 1}=\frac{12 - 3}{-5}=-\frac{9}{5}).

Answer:

(-\frac{9}{5})