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?
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: $\frac{\sqrt{x^4 + y^4}}{\sqrt{x^2 + y^2}} \leq \sqrt{x^2 + y^2} < \delta = \epsilon.$