Question

homework2: problem 13 (1 point) evaluate the limit: $lim_{t
ightarrow - 10}\frac{t^{2}-100}{5t^{2}+60t + 100}=square$ enter dne if the limit does not exist.

Explanation:

Step1: Factor the numerator and denominator

The numerator $t^{2}-100=(t + 10)(t - 10)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$. The denominator $5t^{2}+60t + 100 = 5(t^{2}+12t + 20)=5(t + 10)(t+2)$.
So, $\lim_{t
ightarrow - 10}\frac{t^{2}-100}{5t^{2}+60t + 100}=\lim_{t
ightarrow - 10}\frac{(t + 10)(t - 10)}{5(t + 10)(t + 2)}$.

Step2: Cancel out the common factor

Cancel out the common factor $(t + 10)$ (since $t
eq - 10$ when taking the limit), we get $\lim_{t
ightarrow - 10}\frac{t - 10}{5(t + 2)}$.

Step3: Substitute $t=-10$

Substitute $t=-10$ into $\frac{t - 10}{5(t + 2)}$, we have $\frac{-10-10}{5(-10 + 2)}=\frac{-20}{5\times(-8)}=\frac{-20}{-40}=\frac{1}{2}$.

Answer:

$\frac{1}{2}$