Answer to Question #190057 in Calculus for devy

Question

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

Prove from first principle that

lim f (x, y) = 0.

(x,y)→(2,1)

Answer 1

Given function is-

$f(x,y)=x-2y$

Let $\forall \in >0$ be given

Now consider ,

    $|f(x,y)-f(2,1)| =|x-2y-0|=|x-2y|$

$\Rightarrow |x-2-2(y-1)| \Rightarrow |x-2-2(y-1)|\le |x-2|+2+y-1|$

If we choose $|x-2|<\in, |y-1|<\in$

then $|x-2'-2(y-1)|<\in+2\in=3\in$

   $|x-2-2(y-1)|<3\in=\in$

So, $\in>0, \exist \in >0$ such that

for $|x-2|<\in, |y-1|<\in \Rightarrow f(x,y)-0|<\in$

i.e. $lim_{(x,y)\to (2,1)}f(x,y)=0$