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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.2012-02-21
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    It should be $$\lim_{x \to \infty} \sqrt{\sin{\frac{3}{\sqrt{x}}}} = \sqrt{\sin{3 \sqrt{ \lim_{x \to \infty} \frac{1}{x} }}}$$2012-02-21
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    @sdcvvc, where would the problem break down if the function was not continuous?2012-02-21
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    @sdcvvc, updated the post2012-02-21
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    Oh. I see. Thanks!2012-02-21
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    If $\sin$ were not continuous, then there would be $x_0$ such that $\lim_{x \to x_0} \sin(x) \ne \sin(x_0)$. If you are perverse, you can take this as the definition of "not continuous".2012-02-21
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    And then, even more perverse, you define $f$ to be continuous when it fails to be "not continuous".2012-02-21
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    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 $\left(\sin 3\right)\sqrt{\bullet}$.2012-02-21

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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)