Prove the limit of sinx as x approaches pi/3

[Ryker]

Homework Statement

What is the limit of sinx as x approaches π/3? Prove it.

The Attempt at a Solution

I was wondering, can one approach this using that theorem that reduces the properties of limits of functions to the properties of limits of sequences? Or does that lead to circular logic, because you're saying if x_n tends to π/3, then f(x_n) = tends to sin(π/3) = √3/2, but you haven't actually proven that the sequence that is a sine of another sequence converging to certain number actually converges to the sine of that number.

I was also playing around with trying to prove that the delta-epsilon way, but I haven't gotten anywhere.

So does anyone know how to prove this? Maybe I'm just overthinking this, and it's something really simple, but I guess that's why I'd appreciate your input.

 

[Dick]

Well, of course you have to know SOMETHING about sin to input into the problem. What you generally prove first is the limits of sin(h) and cos(h) as h->0 using geometrical arguments. Now limit sin(x) as x->pi/3 is the same as limit sin(pi/3+h) as h->0. Use a trig addition formula on sin(pi/3+h).

 

[Ryker]

Hmm, well using the trig addition formula you just get lim sin(x) = lim sin(pi/3 + h) = lim sin(pi/3)*cos(h) + sin(h)*cos(pi/3) = lim sin(pi/3). But can you really just say "well that's just √3/2" then? Or is there something else I'm missing?

 

[Dick]

Hmm, well using the trig addition formula you just get lim sin(x) = lim sin(pi/3 + h) = lim sin(pi/3)*cos(h) + sin(h)*cos(pi/3) = lim sin(pi/3). But can you really just say "well that's just √3/2" then? Or is there something else I'm missing?

If you know lim cos(h)=1 and lim sin(h)=0 and you know sin(pi/3)=sqrt(3)/2 and cos(pi/3)=1/2. I don't see much to add to that. Don't you have some theorems about limits of sums and products?

 

[Ryker]

We do, I'm just not sure I could've just made the jump you suggest. Although, come to think of it, sin(pi/3) is then a constant function, independent of h (which is approaching 0), so I guess you can then just say that the limit is √3/2, right? Is that the reason?

Apart from that, another way I think I could prove it is also just saying that we have shown sinx is continuous (which we have), so at all points a lim sinx, as x approaches a, is just f(a) = sin(a). Would that also work?

Oh, and as for the delta-epsilon proof, is it possible to do that in any way?

Thanks.

 

[Dick]

We do, I'm just not sure I could've just made the jump you suggest. Although, come to think of it, sin(pi/3) is then a constant function, independent of h (which is approaching 0), so I guess you can then just say that the limit is √3/2, right? Is that the reason?

Apart from that, another way I think I could prove it is also just saying that we have shown sinx is continuous (which we have), so at all points a lim sinx, as x approaches a, is just f(a) = sin(a). Would that also work?

Oh, and as for the delta-epsilon proof, is it possible to do that in any way?

Thanks.

Mmm. The limit of sin(x) as x->pi/3 is really pretty easy if you've already shown sin(x) is continuous. Just use the definition of continuity. Which you already did. sin(pi/3) being a constant function doesn't help you AT ALL. You don't want lim x->pi/3 sin(pi/3). You want lim x->pi/3 sin(x). And you would only want to resort to a delta-epsilon proof if you hadn't already proved sin(x) is continuous. Which apparently, you did.

 

[Ryker]

If you know lim cos(h)=1 and lim sin(h)=0 and you know sin(pi/3)=sqrt(3)/2 and cos(pi/3)=1/2. I don't see much to add to that.

sin(pi/3) being a constant function doesn't help you AT ALL. You don't want lim x->pi/3 sin(pi/3). You want lim x->pi/3 sin(x).

Wait, but how does then that first thing you suggested, lead to the proof? I'm sorry, I'm really having trouble wrapping my head around this.

And you would only want to resort to a delta-epsilon proof if you hadn't already proved sin(x) is continuous. Which apparently, you did.

Heh, yeah, this one was actually to satisfy my curiosity, as I really want to be able to do it this way, as well, but just can't, either. So if you have any tips here, they would be greatly appreciated.

 

[Dick]

Wait, but how does then that first thing you suggested, lead to the proof? I'm sorry, I'm really having trouble wrapping my head around this. Heh, yeah, this one was actually to satisfy my curiosity, as I really want to be able to do it this way, as well, but just can't, either. So if you have any tips here, they would be greatly appreciated.

Didn't you use a delta-epsilon proof to show sin(x) is continuous to begin with? You might want to just review that proof.

 

[Ryker]

Didn't you use a delta-epsilon proof to show sin(x) is continuous to begin with? You might want to just review that proof.

No, we did it the way you suggested me to tackle the problem at hand, that is with the lim f(x) as x approaches c is equal to lim f(c+h) as h approaches 0.