# Demonstrate that f(x, y) = x^2 + 2y is continuous at (1, 2).

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### 2 Answers

We'll make use of theorem of limits:

`lim(x->1,y->2)f(x,y)=f(1,2)`

Calculate `lim(x->1,y->2)f(x,y), ` so plug in 1 for x and 2 for y:

`lim(x->1,y->2)f(x,y)=1^2+2*2`

`lim(x->1,y->2)f(x,y) = 1+4=5`

Now calculate function f(1,2) so do the same as you did before:

`f(1,2) =1+4=5`

Correlate results:

`lim(x->1,y->2)f(x,y)=5`

`f(1,2) =5`

**Conclusion: The function is continuous at (1,2) because `lim(x->1,y->2)f(x,y)=f(1,2)` **

To prove f(x, y) = x^2 + 2y is continuous at (1, 2)

we choose For |x- 1| < a, |y - 2| < b

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

<= |(x - 1)(x+1)| + 2|y-2| <= [|x-1|^2 - 2|x-1|]+2|y-2|

<=(a^2 - 2a + 2b) = C