Question

find the limit (enter dne if the limit does not exist) hint: rationalize the denominator. $lim_{(x,y)\to(0,0)}\frac{-2x^{2}+2y^{2}}{sqrt{(-2x^{2}+2y^{2}+1) - 1}}$

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $\sqrt{- 2x^{2}+2y^{2}+1}+1$.
\[

$$\begin{align*} &\lim_{(x,y)\to(0,0)}\frac{-2x^{2}+2y^{2}}{\sqrt{-2x^{2}+2y^{2}+1}-1}\times\frac{\sqrt{-2x^{2}+2y^{2}+1}+1}{\sqrt{-2x^{2}+2y^{2}+1}+1}\\ =&\lim_{(x,y)\to(0,0)}\frac{(-2x^{2}+2y^{2})(\sqrt{-2x^{2}+2y^{2}+1}+1)}{(-2x^{2}+2y^{2}+1)-1}\\ =&\lim_{(x,y)\to(0,0)}\frac{(-2x^{2}+2y^{2})(\sqrt{-2x^{2}+2y^{2}+1}+1)}{-2x^{2}+2y^{2}} \end{align*}$$

\]

Step2: Simplify the expression

Cancel out the non - zero factor $-2x^{2}+2y^{2}$ (for $(x,y)
eq(0,0)$).
\[

$$\begin{align*} &\lim_{(x,y)\to(0,0)}(\sqrt{-2x^{2}+2y^{2}+1}+1) \end{align*}$$

\]

Step3: Evaluate the limit

Substitute $x = 0$ and $y = 0$ into the simplified expression.
\[

$$\begin{align*} &\sqrt{-2(0)^{2}+2(0)^{2}+1}+1=\sqrt{1}+1=2 \end{align*}$$

\]

Answer:

$2$