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Nimfa-mama [501]
3 years ago
14

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

Mathematics
1 answer:
gavmur [86]3 years ago
4 0

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)

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