Power limit

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wolly

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[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 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]
 
wolly said:
[math]\lim_{x \to a}\frac{a^{sin(x)}-a^{sin(a)}}{x-a}[/math]
Click to expand...
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?
 
Dr.Peterson said:
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?
Click to expand...
You substitute
[math]u=sin(x)-sin(a)[/math]and in my case
[math]u=x-a[/math]
 
wolly said:
You substitute
[math]u=sin(x)-sin(a)[/math]and in my case
[math]u=x-a[/math]
Click to expand...
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.)
 
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?
wolly said:
Îs there another method?
Click to expand...
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.
 
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
 
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wolly said:
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
Click to expand...
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).

wolly said:
Well what other methods?You didn't show me!
Click to expand...
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?
 
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