Question

attempt 2: 4 attempts remaining. evaluate the following limit. give an exact answer if the limit is a number. otherwise, enter -∞ or ∞ if the limit is infinite, or enter dne if the limit does not exist in another way. \\(\lim_{x \to 8} \left( \frac{x}{x + 8} + \frac{x - 8}{x^2 - 64} \
ight) =\\)

Explanation:

Step1: Factor the denominator

Note that $x^2 - 64 = (x-8)(x+8)$

Step2: Find common denominator

Rewrite the first term to match the common denominator $(x-8)(x+8)$:
$\frac{x}{x+8} = \frac{x(x-8)}{(x+8)(x-8)}$

Step3: Combine the fractions

$$\begin{align*} \frac{x(x-8)}{(x+8)(x-8)} + \frac{x-8}{(x+8)(x-8)}&=\frac{x(x-8)+(x-8)}{(x+8)(x-8)}\\ &=\frac{(x-8)(x+1)}{(x+8)(x-8)} \end{align*}$$

Step4: Cancel common factors

Cancel $(x-8)$ (valid since $x\to8$ means $x
eq8$):
$\frac{x+1}{x+8}$

Step5: Substitute $x=8$

$\frac{8+1}{8+8} = \frac{9}{16}$

Answer:

$\frac{9}{16}$