Question

Find the limit, if it exists. (If an answer does not exist, enter DNE.)

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)