# Finding the limit as $(x,y)\rightarrow (0,0)$ [duplicate]

I know the answer is supposed to be 2 but I do not understand. $${\lim_{(x,y)\rightarrow (0,0)}} \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2} +1}-1}$$ I tried using intuition by reasoning that the numerator will be virtually $0$, and as the value under the root will be slightly greater than $1$, the root of a value slightly greater than $1$ will also be slightly greater than one. Then by subtracting $1$ that means the answer will be $0$.

Please explain where the flaw in my intuition is.

Note that you may let $z=x^2+y^2$. Then, we are asked to find the limit

\begin{align} \lim_{(x,y)\to (0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}&=\lim_{z\to 0}\frac{z}{\sqrt{z+1}-1}\\\\ &=\lim_{z\to 0}\frac{z(\sqrt{z+1}+1)}{z}\\\\ &=2 \end{align}

• I had the same idea. You can arrive at this too by letting $x=r \cos(\theta)$ and $y=r \sin(\theta)$. Then it's simply a problem of finding a limit as $r \to 0$. – Harry Apr 16 '17 at 5:22
• @Harry Polar coordinates works, but is not the way to go here. – Mark Viola Apr 16 '17 at 5:23
• It should be added that my approach does need l'hôpital's rule also to arrive at the correct answer. – Harry Apr 16 '17 at 5:25
• @Harry Certainly your approach is valid. And why are you discussing LHR? It was not used in the solution herein. – Mark Viola Apr 16 '17 at 5:28
• @goldenlinx No need to apologize. Best of luck on the ensuing tests! And you're welcome. My pleasure. -Mark – Mark Viola Apr 16 '17 at 5:50

For this kind of limit problem, first fix $x$ to be 0, and check the limit is 2 when $y$ approaching 0. Then fix $y$ to be 0, and check the limit is 2 when $x$ approaching 0. So you can conclude the limit 2 is correct for this problem.

Above method comes from some old memory when I studied multivariable calculus. If the limit doesn't exist at all, the two trials above will give different values. For detailed example, see "Calculus" by Stewart 6th edition page 908.

• That method is not sufficient - at all. – Mark Viola Apr 16 '17 at 5:20
• That would not be a correct approach- for a limit to exist for this type of situation, the function must approach the same value when the point is approached from $\textit{every}$ path, not just two. – Harry Apr 16 '17 at 5:21
• @Harry Limits are not approached along paths. – Mark Viola Apr 16 '17 at 5:22
• How would it be properly worded? – Harry Apr 16 '17 at 5:24
• For all $\epsilon$, there exists an open disk around $(0,0)$ such that whenever $(x,y)$ is in that disk $\left|\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+1}-1}-2\right|<\epsilon$. The idea of examining a limit along a given set of contours is useful sometimes to show that a limit fails to exist. – Mark Viola Apr 16 '17 at 5:25