Question

Find the limit, if it exists. (If an answer does not exist, enter DNE.)
lim (x, y)→(0, 0) (x^2 + y^2)/((square root x^2 + y^2 + 49) − 7)

Answer 1

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)=14$

Step-by-step explanation:

to find the limit:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\right)$

we need to first rationalize our expression.

$\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\left(\dfrac{\sqrt{x^2+y^2+49}+7}{\sqrt{x^2+y^2+49}+7}\right)$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(\sqrt{x^2+y^2+49}\,)^2-7^2}$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(x^2+y^2)}$

$\sqrt{x^2+y^2+49}+7$

Now this is our simplified expression, we can use our limit now.

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)\\\sqrt{0^2+0^2+49+7}\\7+7\\14$

Limit exists and it is 14 at (0,0)

Answer 2

Rationalizing the denominator, it is found that the result of the limit is of 14.

What is a limit?

A limit is given by the value of function f(x) as x tends to a value.

In this problem, the limit is given by:

$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 49} - 7}$

Applying the standard substitution, we end up with 0/0, hence we have to rationalize the denominator, thus:

$\lim_{(x,y) \rightarrow (0,0)} \frac{x^2 + y^2}{\sqrt{x^2 + y^2 + 49} - 7} \times \frac{\sqrt{x^2 + y^2 + 49} + 7}{\sqrt{x^2 + y^2 + 49} + 7}$

Applying the subtraction of perfect squares:

$\lim_{(x,y) \rightarrow (0,0)} \frac{(x^2 + y^2)(\sqrt{x^2 + y^2 + 49} + 7)}{x^2 + y^2 - 49 + 49}$

We can simplify, hence:

$\lim_{(x,y) \rightarrow (0,0)} \sqrt{x^2 + y^2 + 49} + 7 = \sqrt{49} + 7 = 14$

The result of the limit is of 14.

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