$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 2} $$
$$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 1} \le 0 $$
I have also tried to use the squeeze theorem with two other equations and obtained different values:
$$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 – 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} – 2} $$
$$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} – 1} \le -1 $$
Best Answer
Remember the difference of squares algebraic identity.
$$ A^2 - B^2 = (A - B)(A + B) $$
Why is that useful? With $A = \sqrt{x^2 + y^2 + 1}$ and $B = 1$, the denominator of your expression is $A - B$. With that in mind,
$$ \begin{align} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} &= \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 1} - 1} \cdot \frac{\sqrt{x^2 + y^2 + 1} + 1}{\sqrt{x^2 + y^2 + 1} + 1} \\ &= \frac{(x^2 + y^2)\left(\sqrt{x^2 + y^2 + 1} + 1 \right)}{x^2 + y^2} \\ &= \sqrt{x^2 + y^2 + 1} + 1 \end{align} $$
Now, you can evaluate the limit as $(x, y) \to (0, 0)$ simply by evaluation, since this expression is continuous at the origin.
$$ \lim_{(x, y) \to (0, 0)} \sqrt{x^2 + y^2 + 1} + 1 = \sqrt{0^2 + 0^2 + 1} + 1 = 2. $$