<s>[INST] 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?
 [/INST] 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 $\delta=\varepsilon$.
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