Can I solve this limit algebraically? $$\lim_{x \to 0^-}\frac{{1}}{x^2}-\frac{1}{|x|}.$$
So I know that $|x|$ at $x<0$ is the same as $-(x)$, thus leading to
$$\lim_{x \to 0^-}\frac{{1}}{x^2}+\frac{1}{(x)}.$$
And after a bit of simplification,
$$\lim_{x \to 0^-}\frac{1+x}{x^2}.$$
$$\lim_{x \to 0^-}(1+x)\frac{1}{x^2}.$$
What do I do from here? I know that if you were to input $1/(-0.0001)^2$ you would be approaching $∞$. For $1+x$, you are going to get a value close to 1. But is there a way to solve this algebraically? I'm going to assume the limit of this function is $∞$, but I feel like I am not approaching this properly algebraically.
What do you think?

By convention,
when we find a limit $L$  such $\frac{A}{0}$ with $A\in \mathbb R \cup \{-\infty,+\infty\}$,
we look at the sign of numerator and denominator.


*

*$A>0$ and $0^+$ , $L=+\infty$.

*$A>0$ and $0^-$ , $L=-\infty$.

*$A<0$ and $0^+$, $L=-\infty$.

*$A<0$ and $0^-$ , $L=+\infty$.

