Question

evaluate the limit, if it exists. if not, enter dne below. $lim_{t
ightarrow - 9}\frac{t^{2}-81}{-t^{2}-2t + 63}$ answer =

Explanation:

Step1: Factor the numerator and denominator

The numerator $t^{2}-81$ is a difference - of - squares and can be factored as $(t + 9)(t - 9)$. The denominator $-t^{2}-2t + 63$ can be factored as follows: First, factor out - 1 to get $-(t^{2}+2t - 63)$. Then, factor $t^{2}+2t - 63=(t + 9)(t - 7)$. So the denominator is $-(t + 9)(t - 7)$.
The limit becomes $\lim_{t
ightarrow - 9}\frac{(t + 9)(t - 9)}{-(t + 9)(t - 7)}$.

Step2: Cancel out the common factor

Since $t
ightarrow - 9$, $t
eq - 9$ and we can cancel out the common factor $(t + 9)$ in the numerator and denominator. The limit simplifies to $\lim_{t
ightarrow - 9}\frac{t - 9}{-(t - 7)}$.

Step3: Substitute $t=-9$ into the simplified expression

Substitute $t=-9$ into $\frac{t - 9}{-(t - 7)}$. We get $\frac{-9 - 9}{-(-9 - 7)}=\frac{-18}{-(-16)}=-\frac{9}{8}$.

Answer:

$-\frac{9}{8}$