Question

assignment 3: problem 4 (1 point) let $f(a)=\frac{5}{a - 8}-\frac{80}{a^{2}-64}$. calculate $lim_{a
ightarrow8}f(a)$ by first finding a continuous function which is equal to $f$ everywhere except $a = 8$. $lim_{a
ightarrow8}f(a)=square$

Explanation:

Step1: Factor the denominator

We know that $a^{2}-64=(a - 8)(a + 8)$. So $f(a)=\frac{5}{a - 8}-\frac{80}{(a - 8)(a + 8)}$.

Step2: Find a common - denominator

The common denominator of the two fractions is $(a - 8)(a + 8)$. Then $f(a)=\frac{5(a + 8)-80}{(a - 8)(a + 8)}=\frac{5a+40 - 80}{(a - 8)(a + 8)}=\frac{5a - 40}{(a - 8)(a + 8)}$.

Step3: Simplify the numerator

Factor out 5 from the numerator: $5a - 40 = 5(a - 8)$. So $f(a)=\frac{5(a - 8)}{(a - 8)(a + 8)}$.

Step4: Cancel out the common factor

For $a
eq8$, we can cancel out the $(a - 8)$ terms. Then $f(a)=\frac{5}{a + 8}$ for $a
eq8$.

Step5: Calculate the limit

Now, we find $\lim_{a
ightarrow8}f(a)=\lim_{a
ightarrow8}\frac{5}{a + 8}$. Substitute $a = 8$ into $\frac{5}{a + 8}$, we get $\frac{5}{8 + 8}=\frac{5}{16}$.

Answer:

$\frac{5}{16}$