1. ## Quick Continuity question

Prove that if g:R->R is continuous at a then f(x,y)=g(x) is continuous at (a,b) $\forall b \in R$

So we know
$\forall e>0 \exists d>0 s.t. \forall x \in R$ where |x-a|<d we have |g(x) - g(a)|<e
So I've said as $\forall b \in R$ g(x)=f(x,y) & g(a)=f(a,b), these can be substituted in giving the expression we need except for the condition that $[(x-a)^2 + (y-b)^2]^\frac{1}{2}
This seems to be an incorrect cheat though, am I along the right lines or not?

2. The order of your quantifiers is wrong. In particular, $\forall e>0 \exists d>0 s.t. \forall x \in R$ implies uniform continuity, not continuity...

You seem to have the right answer though; you can use the same epsilon and delta.

3. Originally Posted by kevinlightman
Prove that if g:R->R is continuous at a then f(x,y)=g(x) is continuous at (a,b) $\forall b \in R$

So we know
$\forall e>0 \exists d>0 s.t. \forall x \in R$ where |x-a|<d we have |g(x) - g(a)|<e
So I've said as $\forall b \in R$ g(x)=f(x,y) & g(a)=f(a,b), these can be substituted in giving the expression we need except for the condition that $[(x-a)^2 + (y-b)^2]^\frac{1}{2}
This seems to be an incorrect cheat though, am I along the right lines or not?
Why should that move be an incorrect cheat? And, yes, you are on the right lines. What you need is simply that from $\sqrt{(x-a)^2+(y-b)^2} < \delta$ it follows that $|x-a|<\delta$, which is true: $|x-a| = \sqrt{(x-a)^2}\leq \sqrt{(x-a)^2+(y-b)^2} <\delta$.