limit of quotient

**da kid** · May 5th 2009, 02:51 AM

im supposed to take the limit as (x,y) approaches (0,0) of a function $f(x,y)=\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}$ . pretty sure this is zero. am i right?

**NonCommAlg** · May 5th 2009, 03:35 AM

Originally Posted by da kid

im supposed to take the limit as (x,y) approaches (0,0) of a function $f(x,y)=\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}$ . pretty sure this is zero. am i right?

yes, the limit is 0. given $\epsilon > 0,$ choose $\delta=\epsilon.$ now if $0 < \sqrt{x^2 + y^2} < \delta,$ then: $\frac{\sqrt{x^4 + y^4}}{\sqrt{x^2 + y^2}} \leq \sqrt{x^2 + y^2} < \delta = \epsilon.$

**da kid** · May 5th 2009, 04:03 AM

the confirmation was adequate and appreciated, but you even used the epsilon-delta definition to show it!

that was cool!

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