Question #344285

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


1
Expert's answer
2022-05-24T17:06:18-0400

ANSWER

Let ϵ>0\epsilon >0 , δ=ϵ5\delta=\frac{\epsilon}{5} and (x1)2+(y+1)2<δ\sqrt{(x-1)^{2}+(y+1 )^{2}}<\delta . Since

x1(x1)2+(y+1)2,y+1(x1)2+(y+1)2\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+2y1=(3x3)+(2y+2)3x1+2y(1)<<5δ=5ϵ5=ϵ.|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)(1,1)(3x+2y)=1\lim _{(x,y)\rightarrow(1,-1)}(3x+2y)=1 .


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