Question

find $lim_{x \to 3} \frac{x + 3}{x^2 - 9}$.
enter 1 for $infty$, -1 for $-infty$, and dne if the limit does not exist.
answer:

Explanation:

Step1: Factor the denominator

The denominator $x^2 - 9$ is a difference of squares, so factor it:
$x^2 - 9 = (x-3)(x+3)$

Step2: Simplify the rational function

Cancel the common factor $(x+3)$ (valid for $x
eq -3$, which is true as $x \to 3$):
$\lim_{x \to 3} \frac{x+3}{(x-3)(x+3)} = \lim_{x \to 3} \frac{1}{x-3}$

Step3: Evaluate left-hand limit

As $x \to 3^-$, $x-3 \to 0^-$, so:
$\lim_{x \to 3^-} \frac{1}{x-3} = -\infty$

Step4: Evaluate right-hand limit

As $x \to 3^+$, $x-3 \to 0^+$, so:
$\lim_{x \to 3^+} \frac{1}{x-3} = +\infty$

Step5: Compare one-sided limits

Left-hand limit $
eq$ right-hand limit, so the two-sided limit does not exist.

Answer:

DNE