Question

Find the limit (enter 'DNE' if the limit does not exist)

Answer 1

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)=2$

Step-by-step explanation:

We need to first simplify the expression using rationalization(i.e. if a square root term exists in the denominator, then multiply and divide the whole expression by the denominator(but the change the sign of its middle term))

here, we need to find:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\right)$

first we'll rationalize our expression:

$\dfrac{-2x^2-6y^2}{\sqrt{-2x^2-6y^2+1}-1}\left(\dfrac{\sqrt{-2x^2-6y^2+1}+1}{\sqrt{-2x^2-6y^2+1}+1}\right)$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{(\sqrt{-2x^2-6y^2+1}+1)^2-(1)^2}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-2x^2-6y^2+1-1}$

$\dfrac{-(2x^2+6y^2)(\sqrt{-2x^2-6y^2+1}+1)}{-(2x^2+6y^2)}$

$\sqrt{-2x^2-6y^2+1}+1$

this is our simplified expression, now we can apply our limits:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{-2x^2-6y^2+1}+1\right)$

$\sqrt{-2(0)^2-6(0)^2+1}+1$

$1+1$

$2$

the limit does exists and it is 2.