Question

Find the limit (enter 'DNE' if the limit does not exist)
Hint: rationalize the denominator.

lim(x,y)→(0,0)(−2x^2−6y^2)/sqrt(−2x^2−6y^2+1)-1

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.

Answer 2

Rationalizing the denominator, it is found that limit is of 0.

What is the limit that we want to calculate?

It is given by:

$\lim_{(x,y) \rightarrow (0,0)} \frac{(-2x^2-6y^2)}{\sqrt{(-2x^2-6y^2)} - 1}$

Doing the standard procedure of replacing x and y both by 0, we end up with 0/0, hence we have to rationalize the denominator, that is:

$\lim_{(x,y) \rightarrow (0,0)} \frac{(-2x^2-6y^2)}{\sqrt{(-2x^2-6y^2)} - 1} \times \frac{{\sqrt{(-2x^2-6y^2)} + 1}}{{\sqrt{(-2x^2-6y^2)} + 1}}$

Then, we will have:

$\lim_{(x,y) \rightarrow (0,0)} \frac{(-2x^2-6y^2)(\sqrt{(-2x^2-6y^2)} + 1)}{(-2x^2-6y^2) - 1}$

Hence, replacing, the result is: 0/-1 = 0.

More can be learned about limits at https://brainly.com/question/26270080