Get the answer to "Find the limit, if it exists, or show..."

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Idilwsiw2

Answered question

2021-11-12

Find the limit, if it exists, or show that the limit does not exist. lim (x,y) tends to (0,0)
$\frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1}$

Answer & Explanation

Fearen

Beginner2021-11-13Added 15 answers

What we know
Again the question is using us to evaluate the given limit to see if it exist or not
$lim_{(x, y) \to (0, 0)} \frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1} = ?$
Calculations
We can evaluate the limit by rationalizing the denominator
$lim_{(x, y) \to (0, 0)} \frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1}$
$lim_{(x, y) \to (0, 0)} \frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1} \cdot \frac{\sqrt{x^{2} + y^{2} + 1} + 1}{\sqrt{x^{2} + y^{2} + 1} + 1}$
$lim_{(x, y) \to (0, 0)} \frac{(x^{2} + y^{2}) (\sqrt{x^{2} + y^{2} + 1} + 1)}{x^{2} + y^{2} + 1 - 1}$
$lim_{(x, y) \to (0, 0)} \sqrt{x^{2} + y^{2} + 1} + 1$
$lim_{(x, y) \to (0, 0)} \sqrt{0^{2} + 0^{2} + 1} + 1$
$lim_{(x, y) \to (0, 0)} \frac{x^{2} + y^{2}}{\sqrt{x^{2} + y^{2} + 1} - 1} = 2$

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