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?
 A: Hint :

*

*if $x \rightarrow -5^+$ then $x+5 > 0$


*if $x \rightarrow -5^-$ then $ x+5 < 0$
You'll find that the limit doesn't exist because $\lim_{x\rightarrow -5^+} f(x) \neq \lim_{x\rightarrow -5^-} f(x)$.
A: $$\frac{(x-3)(x+5)}{ |x+5|} = (x-3)\frac{(x+5)}{ |x+5|}=(x-3)\text{sign}(x+5)$$
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.
A: Another place where "factor out the big part" helps.  Here, $|x|$ is growing without bound so $x^2$ is the big part in the numerator and $|x|$ is the big part in the denominator.
\begin{align*}
L &= \lim_{x \rightarrow -\infty} \frac{x^2 + 2x - 15}{|x+5|}   \\
&= \lim_{x \rightarrow -\infty} \frac{x^2(1 + 2/x - 15/x^2)}{|x(1+5/x)|}   \\
    &= \lim_{x \rightarrow -\infty} \frac{x^2}{|x|} \frac{1 + 2/x - 15/x^2}{|1+5/x|}
\end{align*}
We should be able to see that the second fraction is going to $\frac{1+0-0}{|1+0|}$.  (Making this easy to see is why you factor out the "big part".)  The first fraction is always positive (positive over positive) once $x < 0$, so
$$ L = \lim_{x \rightarrow -\infty} (|x| \cdot 1) = \infty  \text{.}  $$
A: After the correction in the limit $x\rightarrow -5$.
Recall that $|x+5|=\begin{cases}x+5 \space\space \text{if $x\geq-5$} \\-(x+5)\space\space \text{if $x<-5$}\end{cases}.$ Then

*

*$\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{-(x+5)}=\lim_{x\rightarrow (-5)^{-}}-(x-3)=-(-5-3)=8$ since we approach from $x<-5$

*$\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{+}}\frac{(x-3)(x+5)}{(x+5)}=\lim_{x\rightarrow (-5)^{+}}(x-3)=-5-3=-8$ since we approach from $x>-5.$
Thus the limit doesn't exist.
A: You even don't have to factor whatever if you use asymptotic equivalents:
a polynomial is equivalent to its leading term, and $|x+5|\sim_{-\infty} |x|$, so
$$\frac{x^2+2x-15 }{|x+5|}\sim_{-\infty} \frac{x^2}{|x|}=\frac{|x|^2}{|x|}=|x|.$$
