# Prove with definition of limits that limit(x^2+2y)=5 if x -->1, y-->2?

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### 1 Answer

`|f(x,y)-5|=|x^2+2y-5|=|x^2-1+2y-4|`

By triangle inequality

`|f(x,y)-5|=|x^2-1+2y-4|<=|x^2-1|+|2y-4|`

`<=|x+1||x-1|+2|y-2|`

as x->1 and y->2 ,we have

`|f(x,y)-5|<=0`

But positive number can not be less than 0

`|f(x,y)-5|=0`

`=>lim_{(x,y)->(1,2)}f(x,y)=5`

where

`f(x,y)=x^2+2y`

Ans.