0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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 $-1$ $\endgroup$ – Andrei Apr 1 '20 at 20:22
  • $\begingroup$ The problem says to prove it using the definition of a limit so just plugging in the values won't work $\endgroup$ – user604720 Apr 1 '20 at 20:23
  • 1
    $\begingroup$ You are not paying attention to my comment. The formula is wrong $\endgroup$ – Andrei Apr 1 '20 at 20:27
0
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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". $\endgroup$ – user604720 Apr 1 '20 at 21:07
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $\delta=\epsilon$.

$\endgroup$
1
  • $\begingroup$ 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". $\endgroup$ – user604720 Apr 1 '20 at 21:07

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.