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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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You can do this whenever the function is known to be continuous. – student Feb 21 '12 at 13:21
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It should be $$\lim_{x \to \infty} \sqrt{\sin{\frac{3}{\sqrt{x}}}} = \sqrt{\sin{3 \sqrt{ \lim_{x \to \infty} \frac{1}{x} }}}$$ – sdcvvc Feb 21 '12 at 13:23
@sdcvvc, where would the problem break down if the function was not continuous? – Inquest Feb 21 '12 at 13:33
@sdcvvc, updated the post – Jiew Meng Feb 21 '12 at 13:36
Oh. I see. Thanks! – Inquest Feb 21 '12 at 13:55
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1 Answer

up vote 2 down vote accepted

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)

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