Question

Consider the R²−R function f defined by f(x,y) = x−2y. Prove from first principles that lim_{(x,y)→(2,1)} f(x,y)=0.

Answer 1

Ans:-

Given that $f(x,y) = x−2y$ which belongs to R²−R function

$\Rightarrow lim_{(x,y)→(2,1)} f(x,y)$

applying first principal

$\Rightarrow lim_{h\to 0}\dfrac{f(x+h,y+h)-f(x)}{h}\\ \Rightarrow lim_{h\to 0}\dfrac{(x+h)-2(y+h)-(x-2y)}{h}\\ \Rightarrow =0$

Hence $lim_{(x,y)→(2,1)} f(x,y)=0$