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How do we solve:

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

Thank You.

Brian M. Scott
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Lavanya
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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 ... . – Brian M. Scott Dec 05 '12 at 05:21

1 Answers1

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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.

Gerry Myerson
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  • Thanks. That works but I'm not sure if we can perform that operation on power functions... Can we? – Lavanya Dec 05 '12 at 05:20
  • Put $ y=\frac{1}{5^x} $ and see what happens. – Mhenni Benghorbal Dec 05 '12 at 05:22
  • Thanks to all of you. I got it now! – Lavanya Dec 05 '12 at 05:31
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    Good. Then you can write it up and post it as an answer. Then, later, you can accept it. – Gerry Myerson Dec 05 '12 at 05:35
  • 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 infinitysin(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 infinity0 form?? – Lavanya Dec 05 '12 at 09:20
  • $\infty\times0$ 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. – Gerry Myerson Dec 05 '12 at 23:49