Limiting Proof: Showing f(v+tej) = b as t→0

[Strangelurker]

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 lim f(v) = b?

 

[ZioX]

Questions:

1. What does the assumed limit tell you?
2. By your logic f(lim t-> 0 v+te_j) = lim t->0 f(v+te_j)=b. Is this true? Why?
3. What mathematical concept are we secretly talking about?

I can only say that intuitive arguments are rarely considered as proofs.

 

[Strangelurker]

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 the standard basis for R^n right?, I guess what I'm mostly trying to ask is that I don't really understand what the f(v+te_j) means (or for that matter taking its limit as t-->0)

 

[ZioX]

You're just taking limits in R^n or whatever.

What is the definition for f being continuous at v?
Is f continuous at v?

So does that mean you can slip in the limit?

 

[Strangelurker]

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 sent to zero we are basically adding nothing to v and taking f(v), I know this last sentence if very sloppy but this is where my logic or lack thereof gets stuck

 

[Dick]

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 sent to zero we are basically adding nothing to v and taking f(v), I know this last sentence if very sloppy but this is where my logic or lack thereof gets stuck

Yeah, that's sloppy. You want to prove lim t->0 of v+t*e_j is v. Do epsilons and deltas. For all epsilon>0, there exists a delta>0 such that |v-(v+t*e_j)|<epsilon if |t|<delta. The relation between epsilon and delta is so painfully obvious, I can't continue. Please help me.

 

[Strangelurker]

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 can get x as close to v as we want, and that te_j will be as small as we want by choosing a t_ne_j, but don't know what to do, sorry if I seem thick this is why I came on here to ask a question was because I am absolutely stuck and need help(it's my first intro to analysis class)

 

[Dick]

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 can get x as close to v as we want, and that te_j will be as small as we want by choosing a t_ne_j, but don't know what to do, sorry if I seem thick this is why I came on here to ask a question was because I am absolutely stuck and need help(it's my first intro to analysis class)

It looks to me like e_j is just supposed to be one of the usual basis vectors for R^n. So |e_j|=1, yes? What's |t*e_j|? I think you are making this harder than it is. There's no 'n' in the limit problem.

 

[HallsofIvy]

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 sent to zero we are basically adding nothing to v and taking f(v), I know this last sentence if very sloppy but this is where my logic or lack thereof gets stuck

There is no reason to believe that f is continuous here. That is not given, nor is it necessary. Also, you don't have to assume that e_j is a basis vector although that is implied by the notation.

If f(x) is any function of a vector argument, continuous at v or not, if \displaytype \lim_{x\to v} f(x)= b implies that \displaytype\lim_{t\to 0} f(v+ tu)= b where u is any fixed vector.

 

[Strangelurker]

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)

 

[Strangelurker]

There is no reason to believe that f is continuous here. That is not given, nor is it necessary. Also, you don't have to assume that e_j is a basis vector although that is implied by the notation.

If f(x) is any function of a vector argument, continuous at v or not, if \displaytype \lim_{x\to v} f(x)= b implies that \displaytype\lim_{t\to 0} f(v+ tu)= b where u is any fixed vector.

How do I show that though?

 

[Dick]

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)?

You were already basically done when you said this, then you went off on some kind of a tangent. You showed delta=epsilon works. So you've shown lim as t->0 of v+t*e_j is v. The problem tells you that lim f(x) as x->v is b. Put v+t*e_j in for x.