How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$$
To do so, I have to use the limit precise definition:
$$\forall\epsilon>0\ \exists\delta>0∶\ \left|\left(x,y\right)-\left(a,b\right)\right|<\delta\ {\Rightarrow}\left|f\left(x,y\right)-L\right|<\ \epsilon$$
I've found out that the the potential limit $L=2$ by evaluating the limits across the x-axis, y-axis, arbitrary line and x^2 parabola and y^2 parabola.
I have plugged all the necessary values and ended up with this:
$$\forall\epsilon>0\ \exists\delta>0∶\ \sqrt{x^2+y^2}<\delta\ {\Rightarrow}\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\ \epsilon$$
Now, I am in trouble to construct inequalities that lead to prove that the following is true:
$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\sqrt{x^2+y^2}\ $$
Can someone help me out here?
 A: You have\begin{align}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}&=\frac{(x^2+y^2)\left(\sqrt{x^2+y^2+1}+1\right)}{\left(\sqrt{x^2+y^2+1}-1\right)\left(\sqrt{x^2+y^2+1}+1\right)}\\&=\sqrt{x^2+y^2+1}+1\end{align}and therefore$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|=\left|\sqrt{x^2+y^2+1}-1\right|.$$But$$\require{cancel}\left|\sqrt{x^2+y^2+1}-1\right|\leqslant\sqrt{x^2+y^2}.$$In fact, this inequality is equivalent to$$\cancel{x^2+y^2}+2-2\sqrt{x^2+y^2+1}\leqslant\cancel{x^2+y^2}$$and it is clear that we always have $2\leqslant2\sqrt{x^2+y^2+1}$.
Putting all together, we have$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|\leqslant\sqrt{x^2+y^2}$$and so, in order to prove that$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=2,$$for each $\varepsilon>0$, we only have to take $\delta=\varepsilon$.
A: A proof involving geometry:
Set $r^2:=x^2+y^2$, where $r \gt 0$.
Need to show: $(r^2+1)^{1/2}-1 \lt r$.
Consider a right triangle with the $2$
legs $r$ and $1$.
Pythagoren theorem:
The hypothenuse is given by $(r^2+1)^{1/2}.$
In any triangle:
The sum of $2$ sides is greater than the $3$rd side.
Hence
$r+1> (r^2+1)^{1/2}$,
$(r^2+1)^{1/2}-1<r$, and we are done.
A: Let $r^2=x^2+y^2$
$|\frac{r^2}{\sqrt{r^2+1}-1}-2|<\epsilon$
$2-\epsilon<\sqrt{r^2+1}+1<2+\epsilon$ by rationalizing the denominator, cancelling terms and pushing $2$ to the outsides of the compound inequality.
$  \sqrt{r^2+1}\le1+\frac{r^2}{2} \implies -\epsilon< r^2/2<\epsilon$ by Taylor's Theorem, then cancelling, then swapping $x$ and $y$ back in.
$x^2+y^2<2\epsilon$. So let  $\delta=\epsilon$ for a simpler, tighter bound.
