Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
\\(lim_{t
ightarrow - 6}\frac{t^{2}-36}{2t^{2}+13t + 6}\\)

Explanation:

Step1: Factor the numerator and denominator

The numerator $t^{2}-36$ is a difference - of - squares and can be factored as $(t + 6)(t - 6)$.
The denominator $2t^{2}+13t + 6$ can be factored by splitting the middle - term. We need to find two numbers that multiply to $2\times6 = 12$ and add up to $13$. The numbers are $12$ and $1$. So, $2t^{2}+13t + 6=2t^{2}+12t+t + 6=2t(t + 6)+1(t + 6)=(2t + 1)(t + 6)$.
So, the function becomes $\lim_{t
ightarrow - 6}\frac{(t + 6)(t - 6)}{(2t + 1)(t + 6)}$.

Step2: Cancel out the common factor

Since $t
eq - 6$ (we are taking the limit as $t$ approaches $-6$), we can cancel out the common factor $(t + 6)$ in the numerator and denominator. The function simplifies to $\lim_{t
ightarrow - 6}\frac{t - 6}{2t+1}$.

Step3: Substitute $t=-6$ into the simplified function

Substitute $t=-6$ into $\frac{t - 6}{2t + 1}$. We get $\frac{-6-6}{2\times(-6)+1}=\frac{-12}{-12 + 1}=\frac{-12}{-11}=\frac{12}{11}$.

Answer:

$\frac{12}{11}$