Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?

$$\frac{(x-3)(x+5)}{ |x+5|} = (x-3)\frac{(x+5)}{ |x+5|}=(x-3)\text{sign}(x+5)$$
As $x \to -\infty$, then $\text{sign}(x+5)=-1$, So $(x-3)\text{sign}(x+5)=-x+3 \to \infty$.
Addition.
If we take limit point $x=-5$, then of course we come from infinite limit to non existence of limit. First multiplier $x-3 \to -8, x \to -5$, but second as $\text{sign}(x+5)$ have different one-sided limits $\pm 1.$ So $\mp 8$ for product.
