im supposed to take the limit as (x,y) approaches (0,0) of a function $\displaystyle f(x,y)=\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}$. pretty sure this is zero. am i right?
im supposed to take the limit as (x,y) approaches (0,0) of a function $\displaystyle f(x,y)=\frac{\sqrt{x^4+y^4}}{\sqrt{x^2+y^2}}$. pretty sure this is zero. am i right?
yes, the limit is 0. given $\displaystyle \epsilon > 0,$ choose $\displaystyle \delta=\epsilon.$ now if $\displaystyle 0 < \sqrt{x^2 + y^2} < \delta,$ then: $\displaystyle \frac{\sqrt{x^4 + y^4}}{\sqrt{x^2 + y^2}} \leq \sqrt{x^2 + y^2} < \delta = \epsilon.$