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?


2 Answers 2


You defined $u$ by $u=\sqrt{x}>0$. So when $x=4$ we have $u=\sqrt{4}=2$.

  • $\begingroup$ Ah, that makes sense. Silly oversight on my part. $\endgroup$
    – Jack Pan
    Commented Jul 22, 2016 at 14:30

Perhaps it should be made explicit verbally at the outset that the chosen substitution is that $u$ is the positive square root of $x$. In the problem as stated before the substitution is done, the numerator is $\sqrt x - 2$, and $\text{“}\sqrt x\text{ ''}$ denotes one of the two square roots of $x$ and not the other one.

If what one knows about $u$ is ONLY that $u^2$ is approaching $4$, then one might conclude in certain contexts that $u$ is approaching either $2$ or $-2$ but one cannot tell which. But in a broader context even that conclusion is not justified: $u$ could be alternating between something approaching $2$ and something approaching $-2$, as $u^2$ approaches $4$, so that $u$ would not be approaching any limit at all. However, in this case the “ONLY” mentioned above is not applicable.


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