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
Let ϵ>0\epsilon >0ϵ>0 , δ=ϵ5\delta=\frac{\epsilon}{5}δ=5ϵ and (x−1)2+(y+1)2<δ\sqrt{(x-1)^{2}+(y+1 )^{2}}<\delta(x−1)2+(y+1)2<δ . Since
∣x−1∣≤(x−1)2+(y+1)2,∣y+1∣≤(x−1)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}}∣x−1∣≤(x−1)2+(y+1)2,∣y+1∣≤(x−1)2+(y+1)2 , then ∣f(x,y)−1∣=∣3x+2y−1∣=∣(3x−3)+(2y+2)∣≤3∣x−1∣+2∣y−(−1)∣<<5δ=5⋅ϵ5=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.∣f(x,y)−1∣=∣3x+2y−1∣=∣(3x−3)+(2y+2)∣≤3∣x−1∣+2∣y−(−1)∣<<5δ=5⋅5ϵ=ϵ.
So, lim(x,y)→(1,−1)(3x+2y)=1\lim _{(x,y)\rightarrow(1,-1)}(3x+2y)=1lim(x,y)→(1,−1)(3x+2y)=1 .
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