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)`**

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