Answer to Question #344285 in Calculus for Sexyeyes

Question

Consider the R^2 - R function defined by f(x,y) = 3x + 2y.

Prove from first principles that lim (x,y) -> (1,-1) f(x,y) = 1

Answer 1

ANSWER

Let $\epsilon >0$ , $\delta=\frac{\epsilon}{5}$ and $\sqrt{(x-1)^{2}+(y+1 )^{2}}<\delta$ . Since

$\left | x-1 \right |\leq \sqrt{(x-1)^{2}+(y+1 )^{2}}, \left | y+1 \right |\leq \sqrt{(x-1)^{2}+(y+1 )^{2}}$ , then $|f(x,y)-1|=|3x+2y-1|=|(3x-3 )+(2y+2)|\leq 3|x-1|+2|y-(-1)|<\\<5\delta=5\cdot \frac {\epsilon}{5}=\epsilon.$

So, $\lim _{(x,y)\rightarrow(1,-1)}(3x+2y)=1$ .