Thread: Limits in R^n

This may be a silly problem to be having but i've had a couple of year off from uni maths and have forgotten some things i should know but aren't covered in current courseany help would be greatly appreciated cheers,

Question i'm doing off a practice sheet;

Let f:$\displaystyle R^2$--> R be defined by f(x1,x2) = (x1)^2 + (x2)^2
Prove that, lim(x->0) f(x) = 0.

The difficulty i'm having is with the mapping.. i realise that the function maps 2 dimensional vector space to 1d but i'm having trouble understanding the effects this has on the question, i'm good with the epsilon delta in the actual finding of limit but don't think i understand the mapping of the function correctly.. any help would be greatly appreciated cheers.

to be more specific in the asnwers it says; (where E = epsilon)

l x1^2 + x2^2 - 0 l < E (i understand getting to here)

therefore
x1^2 + x2^2 < E

therefore ll (x1,x2) ll^2 < E

but not he next two steps,

any help would be appreciated,
cheers

The definition that a function g : R -> R of one argument is continuous at 0 (which is the same thing as $\displaystyle \lim_{x\to0}g(x)=g(0)$) is:

For all ε > 0 there exists a δ > 0 such that for any x, if |x| < δ, then |g(x) - g(0)| < ε.

Here |x| can be viewed as the magnitude of x, or the norm of one-dimensional vector x. Note also that $\displaystyle |x| = \sqrt{x^2}$.

When f : R x R -> R, we consider a similar norm on two-dimensional argument vectors: $\displaystyle \|(x_1,x_2)\| = \sqrt{x_1^2 + x_2^2}$. The definition of continuity of f at 0 stays the same:

For all ε > 0 there exists a δ > 0 such that for any (x1,x2), if ||(x1,x2)|| < δ, then |f(x1,x2) - f(0,0)| < ε.

To prove this, fix any ε > 0 and let δ = sqrt(ε). Take any (x1,x2) and suppose that ||(x1,x2)|| < δ, i.e., $\displaystyle \sqrt{x_1^2+x_2^2}<\sqrt{\varepsilon}$. Then $\displaystyle 0\le x_1^2+x_2^2=f(x_1,x_2)=f(x_1,x_2)-f(0,0)<\varepsilon$, so $\displaystyle |f(x_1,x_2)-f(0,0)|<\varepsilon$, as required.

It is also possible to choose δ = min(ε,1) because then δ^2 <= δ <= ε.

to be more specific in the asnwers it says; (where E = epsilon)

l x1^2 + x2^2 - 0 l < E (i understand getting to here)

therefore
x1^2 + x2^2 < E

Here you start with what you need to prove, so it's not right to say "therefore."