Limites | Mathematics | Page 1

dire ruin Apr 2, 2024, 2:33 PM

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please need help for 13

regal cometBOT Apr 2, 2024, 2:33 PM

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hasty vector Apr 2, 2024, 2:44 PM

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dire ruin please need help for 13

Divide both numerator and denominator of #13 by xy (since neither x nor y can be 0), get ±1/sqrt(1/x^2 + 1/y^2), where plus corresponds to xy being > 0, minus to xy being < 0. Now the denominator is clearly unbounded, so the answer is 0.

dire ruin Apr 2, 2024, 3:01 PM

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I am sorry but I didn’t get you explanation

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@hasty vector

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is this way good ?

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