Give a rigorus epsilon-delta proof that the function f: R^3 into R defined by f(x,y,z) = yx^2 + 2xz^2 is continuous at (1,1,1).
suppose $\displaystyle \epsilon > 0$ is given and choose $\displaystyle \delta = \min \{1, \epsilon/21 \}.$ now if $\displaystyle 0 < |x-1|, |y-1|, |z-1| < \delta,$ then $\displaystyle 0 <x,y,z < 2$ because $\displaystyle \delta \leq 1.$
we also have: $\displaystyle yx^2+2xz^2 - 3 = (y-1)x^2 + (x-1)(x+1+2z^2) + 2(z-1)(z+1).$ thus: $\displaystyle |yx^2+2xz^2-3| < 4 \delta + 11 \delta + 6 \delta = 21 \delta < \epsilon.$