# Proving a limit with 2 variables using definition

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.

• I think your first equation is wrong. Just plug in $x=y=0$ and you get $0/2=0$. My guess is that the last $+1$ should be instead $-1$ – Andrei Apr 1 '20 at 20:22
• The problem says to prove it using the definition of a limit so just plugging in the values won't work – user604720 Apr 1 '20 at 20:23
• You are not paying attention to my comment. The formula is wrong – Andrei Apr 1 '20 at 20:27

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 $$\sqrt{x^2+y^2} < \delta \implies |f(x,y)| \le x^2+y^2 < \delta^2 = \epsilon$$.

• I am sorry, but I made an error typing the problem. I fixed the problem by changing the "+1" in the denominator to a "-1". – user604720 Apr 1 '20 at 21:07

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

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

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 $$\delta=\epsilon$$.

• I am sorry, but I made an error typing the problem. I fixed the problem by changing the "+1" in the denominator to a "-1". – user604720 Apr 1 '20 at 21:07