5
$\begingroup$

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?

$\endgroup$
10
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Ah, that makes sense. Silly oversight on my part. $\endgroup$ – Max Li Jul 22 '16 at 14:30
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.