Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne) $lim_{t
ightarrow7}\frac{t^{2}-t - 42}{t - 7}$

Explanation:

Step1: Factor the numerator

We factor $t^{2}-t - 42$ as $(t - 7)(t+6)$ using the formula $ax^{2}+bx + c=a(x - x_1)(x - x_2)$ where $x_1,x_2$ are roots of $ax^{2}+bx + c = 0$. For $t^{2}-t - 42=0$, the roots are $t=\frac{1\pm\sqrt{1+168}}{2}=\frac{1\pm13}{2}$, so $t = 7$ and $t=-6$. So $\lim_{t
ightarrow7}\frac{t^{2}-t - 42}{t - 7}=\lim_{t
ightarrow7}\frac{(t - 7)(t + 6)}{t - 7}$.

Step2: Cancel out the common factor

Since $t
eq7$ when taking the limit (we are considering values of $t$ approaching 7 but not equal to 7), we can cancel out the $(t - 7)$ terms. So $\lim_{t
ightarrow7}\frac{(t - 7)(t + 6)}{t - 7}=\lim_{t
ightarrow7}(t + 6)$.

Step3: Evaluate the limit

Substitute $t = 7$ into $t + 6$. We get $\lim_{t
ightarrow7}(t + 6)=7+6=13$.

Answer:

$13$