Proving a limit with 2 variables using definition

Question

Use the definition of a limit to show that $$\lim_{(x,y)\to (0,0)}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}=2$$ In other words, show that for every real number $\epsilon>0$ you can find a real number $\delta>0$ such that whenever $\sqrt{(x-0)^2+(y-0)^2}<\delta$ then $|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}+1}-2|<\epsilon$

What I did to simplify the fraction is I multiplied it by it's conjugate and then simplified to get: $\sqrt{x^2+y^2+1}+1$

However, I am not quite sure where to go from here.

Answer 1

The problem is easier than you think.The limit value on the right should be $0$, and it's just a minor error. Observe that $\sqrt{x^2+y^2+1}+1 > 1$ for all $x,y$. Thus if you take $\delta = \sqrt{\epsilon}$, then the conclusion follows since $\sqrt{x^2+y^2} < \delta \implies |f(x,y)| \le x^2+y^2 < \delta^2 = \epsilon$.

Answer 2

The limit is $0$.

$|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}+1}-0| \le$

$\frac{x^2+y^2}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2};$

Choose $\delta =\epsilon$.

The re-edited problem:

Set $t:=x^2+y^2$.

Show that

$\lim_{t \rightarrow 0^+}\frac{t}{\sqrt{t+1}-1}=2$;

$\frac {t}{\sqrt{t+1}-1}=\frac{t(\sqrt{t+1}+1)}{t}=$

$=\sqrt{t+1}+1$;

$|\sqrt{t+1}+1-2|= |\sqrt{t+1}-1|=$

$|\frac{t}{\sqrt{t+1}+1}|\lt t.$

Choose $\delta=\epsilon$.

Answer 3

Let $x^2+y^2+1=z^2$. Then $\sqrt{x^2+y^2+1}=\pm z$.

If Andrei's correction is made then taking the +ve option $$\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{z\to +1}\frac{z^2-1}{z-1}= \lim_{z \to +1} (z+1) = +2$$

while taking the -ve option $$\lim_{(x,y)\to (0,0)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=\lim_{z\to -1}\frac{z^2-1}{-z-1}= \lim_{z \to -1} -(z-1) = +2$$

Answer 4

I will show you the corrected problem: $$\lim_{(x,y)\to (0,0)}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}\color{red}-1}}=2$$ Multiply with the conjugate and you get $$\begin{align}\lim_{(x,y)\to (0,0)}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}&=\lim_{(x,y)\to (0,0)}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}\frac{\sqrt{x^2+y^2+1}+1}{\sqrt{x^2+y^2+1}+1}\\&=\lim_{(x,y)\to (0,0)}{\frac{(x^2+y^2)(\sqrt{x^2+y^2+1}+1)}{x^2+y^2+1-1}}\\&=\lim_{(x,y)\to (0,0)}(\sqrt{x^2+y^2+1}+1)\end{align}$$ Now you can use the $\varepsilon-\delta$ definition