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?

For $x\lt 0$ one has $f(x)=\dfrac{1}{x^2}-\dfrac{1}{|x|}=\dfrac{1}{x^2}-\dfrac{1}{-x}=\dfrac{1}{x^2}+\dfrac{1}{x}=$.
For $x\gt 0$ one has $f(x)=\dfrac{1}{x^2}-\dfrac{1}{|x|}=\dfrac{1}{x^2}-\dfrac{1}{x}$ It follows
$$\begin{cases}\lim_{x\to 0^{-}}f(x)=\infty\\\lim_{x\to 0^{+}}f(x)=\text{ indeterminated form }\space \infty-\infty\end{cases}$$
However $$\dfrac{1}{x^2}-\dfrac{1}{x}=\frac{x-x^2}{x^3}$$ which is of the form $\dfrac00$ so we can apply l'Hôpital rule so we get the final limit being equal to $\infty$ because it is the limit $$\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}\frac{-2}{6x}=\infty.$$
