Say I have $$\lim_{x \rightarrow 4} f(x)=\frac{\sqrt{x}-2}{\sqrt{x^3}-8}.$$
My homework paper says to do a change of variable for $u=\sqrt{x}.$ If I do that, I get
$$\lim_{u^2 \rightarrow 4} f(x)=\frac{u-2}{\sqrt{u^6}-8},$$
and from there we simplify like the following:
$$\lim_{u^2 \rightarrow 4} f(x)=\frac{u-2}{u^3-8}$$
$$\lim_{u^2 \rightarrow 4} f(x)=\frac{u-2}{(u-2)(u^2+2u+4)}$$
$$\lim_{u^2 \rightarrow 4} f(x)=\frac{1}{u^2+2u+4}$$
My question is, when I square root the $u^2$ as in $u^2\rightarrow4$, does it become $u\rightarrow2$ or $u\rightarrow-2$, and why? It's apparent that in the from the first equation that it "makes sense" or at least "follows some pattern" for $u$ to equal positive $2$ because it would make the numerator $0,$ but this isn't really valid reasoning.
So, should $u$ approach $2$ or $-2,$ and how do we know?
