In my maths lecture notes:
$$\lim_{x \to \infty} \sqrt{\sin{\frac{3}{\sqrt{x}}}} = \sqrt{\sin{3 \sqrt{ \lim_{x \to \infty} \frac{1}{x} }}}$$
When can I move the $\lim$ into a function like this?
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3$\begingroup$ You can do this whenever the function is known to be continuous. $\endgroup$– studentFeb 21, 2012 at 13:21
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2$\begingroup$ It should be $$\lim_{x \to \infty} \sqrt{\sin{\frac{3}{\sqrt{x}}}} = \sqrt{\sin{3 \sqrt{ \lim_{x \to \infty} \frac{1}{x} }}}$$ $\endgroup$– sdcvvcFeb 21, 2012 at 13:23
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$\begingroup$ @sdcvvc, where would the problem break down if the function was not continuous? $\endgroup$– InquestFeb 21, 2012 at 13:33
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$\begingroup$ @sdcvvc, updated the post $\endgroup$– Jiew MengFeb 21, 2012 at 13:36
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1$\begingroup$ I'd prefer to write $\sin\left(3\sqrt{\bullet}\right)$ rather than $\sin 3\sqrt{\bullet}$, to make sure it wouldn't be mistaken for $\left(\sin 3\right)\sqrt{\bullet}$. $\endgroup$– Michael HardyFeb 21, 2012 at 16:06
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1 Answer
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Yes. Here is one (non-rigor) method of looking at it.
Let $\frac{1}{ x} = t$
As ${x \to \infty}, t \to 0 $
$$\lim_{x \to \infty} \sqrt{\sin{\frac{3}{\sqrt{x}}}} = \lim_{t \to 0} \sqrt{\sin (3\sqrt{t})} = \sqrt{\sin(3 \times \lim_{t \to 0} \sqrt{t})}$$ (Owing to Continuity)
