Recent content by Strangelurker

How do I show that though?

Yes you're right, e_j is just the standard basis for j=1,2,...n for R^n, so |t*e_j| is just |t| then, which --> 0, I can describe it clearly in english as distances, but I just can't get down the answer on paper (I know this is the whole point)

So I have that part, that epsilon = delta works for if |t|< delta then |te_j|<epsilon, so the lim as t--> 0 of (v+te_j) = v, so then the lim t-->0 of f(v+te_j)=f(v)? We know that we can choose N s.t. |x-v|<epsilon for all n>=N, but I don't know how to get this part into the f(v+te_j), I know we...

One equivalent def. for f being continuous at v is if lim x-->v of f(x) = f(v), I don't see how we could assume continuity without an explicit function or more information... So basically we have the vector v + the standard basis in R^whatever with a t multiplying the standard basis. As t gets...

1. Besides the epsilon/delta definition? That if x_n --> v, then f(x_n)--> v? 2. This would be true if we had continuity right? The problem does not say so therefore I'm guessing that's not it. I don't know why I'm getting so stuck/frustrated with this problem... the te_j for j= 1,2,...,n is...

Homework Statement The problem is if the limit of f(x) as x---> v = b, then show the limit of f(v+ tej) = b as t->0, should read e subscript j Homework Equations The Attempt at a Solution Can you just say that as t-->0, te(subscript j) goes to 0 so by using epsilon/delta argument...