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How do we solve:

$$\lim_{x\to \infty} 5^x \sin\left(\frac{a}{5^x}\right)$$

Thank You.

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    Gerry Myerson’s answer is the way to go, but you can easily see what the limit has to be if you remember that $\sin x\approx x$ when $|x|$ is small. Thus, $\sin\frac{a}{5^x}\approx\frac{a}{5^x}$ when $x$ is large, and ... .2012-12-05

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Convince yourself that it's the same as evaluating $\lim_{t\to0}{\sin at\over t}$, and then use other stuff you know to do that one.

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    Thanks. That works but I'm not sure if we can perform that operation on power functions... Can we?2012-12-05
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    Put $ y=\frac{1}{5^x} $ and see what happens.2012-12-05
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    Thanks to all of you. I got it now!2012-12-05
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    Good. Then you can write it up and post it as an answer. Then, later, you can accept it.2012-12-05
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    Well, now I have got another doubt.. Instead of doing it the above way, if I let it remain x -> infinity, then 5^infinity becomes infinity...so on solving I would get infinity*sin(0) which is all too confusing! And if I can't directly put infinity in place of x, why is it so? Because I'll get an infinity*0 form??2012-12-05
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    $\infty\times0$ is what is known as an indeterminate form. So are $0/0,\infty-\infty,1^{\infty},0^0$, maybe a few others I'm forgetting. If you have a formula $f(x)$, and if $\lim_{x\to Q}f(x)$ results in an indeterminate form, then you have to do some extra work to evaluate that limit. The extra work may take the form of some algebraic manipulation, or some reasoning from geometry, or (once you get to Calculus) l'Hopital's Rule, depending on the exact form of the formula $f$. You have much to look forward to.2012-12-05