Power limit

[wolly]

[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x^2-a^2}[/math]
Do I use the limit of

[math]\lim_{x \to 0}\frac{a^{x}-1}{x}[/math]?

I am really stuck here!

 

[wolly]

I tried to solve this limit and I got from

[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x^2-a^2}[/math]=[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{(x-a)(x+a)}[/math]to
[math]lim_{x \to a}\frac{1}{x+a}[/math]
[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x-a}[/math]

 

[Dr.Peterson]

[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x-a}[/math]

What would you do if that limit had been [math]\lim_{x \to a}\frac{a^{\sin(x)}-a^{\sin(a)}}{\sin(x)-\sin(a)}[/math]?

Could you use that as part of your work?

 

[wolly]

What would you do if that limit had been [math]\lim_{x \to a}\frac{a^{\sin(x)}-a^{\sin(a)}}{\sin(x)-\sin(a)}[/math]?

Could you use that as part of your work?

You substitute
[math]u=sin(x)-sin(a)[/math]and in my case
[math]u=x-a[/math]

 

[Dr.Peterson]

You substitute
[math]u=sin(x)-sin(a)[/math]and in my case
[math]u=x-a[/math]

I don't see how that helps. What would you actually do?

I'm hoping to see some more actual work, either on your problem, or on my problem, taken on its own. (I think the ideas from it can then be applied to yours.)

 

[wolly]

 

[wolly]

 

[wolly]

Îs there another method?
@Dr.Peterson

 

[Dr.Peterson]

I'm having trouble following what you wrote. Can you explain, for example, the second line, in words? And then, how you replace sin(u) with u and cos(u) with 1, on the fourth line?

Îs there another method?

Yes, there are usually multiple methods!

Have you told us yet whether you can use L'Hopital's rule? That's probably the first thing I'd do, coming at this on my own.

 

[wolly]

Well I used the limit of sinu/u
[math]\lim_{u \to 0}\frac{sin(u)}{u}[/math]and after that I used this part
[math]\lim_{u \to 0}\frac{sin(u)}{u}*u=sin(u)[/math]And sinu/u îs 1 and 1*u îs u
@Dr.Peterson

 

[wolly]

Yes, there are usually multiple methods!

Well what other methods?You didn't show me!

 

[wolly]

I calculated without these parts because I was hurried

 

[wolly]

Well cos(u)=cos(0)=1
I wrote that u tends to 0 in the image

 

[Dr.Peterson]

Well I used the limit of sinu/u
[math]\lim_{u \to 0}\frac{sin(u)}{u}[/math]and after that I used this part
[math]\lim_{u \to 0}\frac{sin(u)}{u}*u=sin(u)[/math]And sinu/u îs 1 and 1*u îs u
@Dr.Peterson

No, you can't let the variable go to its limit in part of an expression, but not in others. This isn't a valid method (except where a theorem allows it, as when you break an expression into a product and take the limit on each factor).

Well what other methods?You didn't show me!

My point is that you can expect that there will be more than one way to solve any problem. I don't have a specific list of all ways to solve this problem.

You haven't answered my question about l'Hopital's rule; if you know it, then we could discuss that. (It leads very easily to a rather ugly answer.)

If you want to use the known limit you mentioned in the OP, try letting [imath]u=\sin(x)-\sin(a)[/imath]. Do you see how you could set things up so that could be useful?

 

[mario99]

[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x^2-a^2}[/math]
Do I use the limit of

[math]\lim_{x \to 0}\frac{a^{x}-1}{x}[/math]?

I am really stuck here!

I have a way to solve the problem using the identity:

[imath]\displaystyle \lim_{x\rightarrow 0}\frac{a^x - 1}{x} = \log a[/imath]

I don't know if professor Dave has figured out that yet.

Have you told us yet whether you can use L'Hopital's rule? That's probably the first thing I'd do, coming at this on my own.

Well I think that the OP wanna be fancy and challenge himself to not solve it by the boring way (L'hopital rule).

 

[Dr.Peterson]

I don't know if professor Dave has figured out that yet.

Yes, I have; my work by that approach includes the substitution I suggested as a key step, as I indicated here:

If you want to use the known limit you mentioned in the OP, try letting [imath]u=\sin(x)-\sin(a)[/imath]. Do you see how you could set things up so that could be useful?

Well I think that the OP wanna be fancy and challenge himself to not solve it by the boring way (L'hopital rule).

It's entirely possible that it has not been taught yet. Limits like the one you cite could even be taught before introducing derivatives at all (and later used to prove that derivative), though I don't think any of the books I have put it in that order. I try to avoid assuming that knowledge when the work shown suggests otherwise.

 

[mario99]

Yes, I have; my work by that approach includes the substitution I suggested as a key step, as I indicated here:



It's entirely possible that it has not been taught yet. Limits like the one you cite could even be taught before introducing derivatives at all (and later used to prove that derivative), though I don't think any of the books I have put it in that order. I try to avoid assuming that knowledge when the work shown suggests otherwise.

I know the style of the OP. He is someone who is going to ignore the whole thread if he did not understand your hint. Therefore, I suggest that you show him the first few steps.

Any way my approach is what you should do if your limit does not match the identity exactly.

First, you have to notice this:

[imath]\displaystyle \lim_{x\rightarrow 0}\frac{a^x - 1}{x} = \lim_{x\rightarrow a}\frac{a^{x - a} - 1}{x - a} = \log a[/imath]

Then, you introduce [imath]A(x,a)[/imath] and [imath]B(x,a)[/imath]:

[imath]\displaystyle \lim_{x\rightarrow a}\frac{a^{A(x,a)} - 1}{B(x,a)} = C(a)\log a[/imath]

where [imath]C(a)[/imath] is a secret function for now.

And the main point of doing that is how to find [imath]C(a)[/imath] when [imath]\displaystyle A(x,a) \neq B(x,a)[/imath] as in the OP limit. It is a beautiful idea, but it has some restrictions.