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I have been solving a 3D limit problem: $$\lim_{(x,y) \to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$which apparently, is $2$. However, I cannot find the flaw in my approach. Here it goes: $$\mathrm{Let \space y=mx} \\\lim_{(x,y) \to (0,0)} \frac{x^2+m^2x^2}{\sqrt{x^2+m^2x^2+1}-1}= \lim_{(x,y) \to (0,0)} \frac{x^2(1+m^2)}{\sqrt{2x^2+1}-1}=\frac{0}{0} \\ \mathrm{Apply \space L'Hôpital's \space rule}\\ \lim_{(x,y) \to (0,0)}\frac{2x(1+m^2)}{(\frac{2x}{\sqrt{2x^2+1}})}$$ I've solved this two different ways and I keep getting the answer as $(1+m^2)$. The Student Solutions Manual, alas, says the answer it $2$. Just in case my mistake lies somewhere after this step, even though I got the same answer solving it two different ways after this point I'll give you one of my methods, the shorter of the two: $$...\space=\lim_{(x,y) \to (0,0)} \frac{2x(1+m^2)\sqrt{2x^2+1}}{2x}=\lim_{(x,y) \to (0,0)}\frac{(1+m^2)\sqrt{2x^2+1}}{1}=(1+m^2)$$ No idea how to get to $2$ here. Also, am I abusing notation by leaving the bounds of the limits in terms of $(x,y)$ rather than reducing it to $x$ when I have let $y$ be equal to some terms of $x$?

user1833028
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5 Answers5

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Hint: Use polar coordinates, and rationalize the denominator. (When one sees $x^2+y^2$, which is often, switching to polar coordinates is usually the right approach.)

In this case, we do not even have to use polar coordinates, rationalizing the denominator is enough.

Remark: Using $y=mx$ is probably not a good start, since given that $(x,y)\to(0,0)$ gives us no useful information about their ratio.

André Nicolas
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  • An addendum to the last comment: it is possible in theory for $\lim_{x\to0} f(x,mx) = c$ for every $m$, yet $\lim_{(x,y)\to(0,0)} f(x,y)$ to not exist. Perhaps $\lim_{x\to0} f(x,x^2) = -c$, for example. (I see this is what @sahimat is saying in his/her answer.) – Greg Martin Oct 07 '14 at 22:06
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Rationalize the denominator:

\begin{equation} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}\cdot \frac{\sqrt{x^2+y^2+1}+1}{\sqrt{x^2+y^2+1}+1} \end{equation}

\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{(x^2+y^2+\sqrt{x^2+y^2+1}-\sqrt{x^2+y^2+1}+1)-1} \end{equation}

\begin{equation}=\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{x^2+y^2} \end{equation}

\begin{equation}=\sqrt{x^2+y^2+1}+1 \, . \end{equation}

M A Pelto
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All of these answers are good, I personally always rationalize the denominator whenever I see something like that as a first try.

You should also remember that you cannot prove a limit exists by using the approach method. You can show a limit does NOT exist by showing it does not yield the same result from 2 different approaches (be that different, lines, parabolas, trig fns etc).

Showing a limit exists you'll want to: 1. manipulate it using algebra (like rationalizing the denominator here), 2. using polar/spherical coordinates (in the case of R2 or R3) and showing as the radius goes to 0, or 3. squeeze theorem, which is easiest when you're given something going to (0), where you can use the property that the if the limit of the absolute value of a function goes to 0, then the function itself will also go to 0 (again, used for manipulating squeeze theorem).

sahimat
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Try using polar coordinates:

$$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\;\;\;\;\implies\;\;\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}\rightarrow\frac{r^2}{\sqrt{r^2+1}-1}=\frac{r^2\left(\sqrt{r^2+1}+1\right)}{r^2}=$$

$$=\sqrt{r^2+1}+1\xrightarrow[r\to 0^+]{}2$$

Timbuc
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If you set $\ x=\rho\cos(\theta), y=\rho\sin(\theta)$ , you have: $$\lim_{\rho\to0}\frac{\rho^2}{\sqrt{1+\rho^2}-1}= \lim_{\rho\to0} \frac{\rho^2}{1+\frac{\rho^2}{2}+o(\rho^2)-1}=\frac{1}{\frac{1}{2}}=2 $$

Where I used the fact that $\sqrt{1+x}=1+\frac{x}{2}+o(x)$ , for $\ x\to0$

Mosk
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