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Sorry for the vague title, I feel like I cannot explain it properly with such few words, so I'll try here:

$$\lim_{x\rightarrow0^+}\left( \frac{x\arctan{(1/x)}}{\sin{3x}} \right) = \lim_{x\rightarrow0^+}\left( \frac{x\cdot\frac{\pi}{2}}{\sin{3x}} \right)$$

Would evaluating first $\arctan({1/x})$ and then use L'hopital's rule be allowed? Or would I have to use L'hopital's rule first before evaluating any factor? I still seem to get the right answer though, which is $\frac{\pi}{6}$.

Freeze Aim
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    What you're using is an application of the following rule: the limit of a product is the product of limits. Do you see how that applies here? – Sambo Oct 27 '22 at 20:39
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    The rule mentioned by @Sambo is true only if all the limits exist and are finite – Andrei Oct 27 '22 at 20:41
  • @Sambo I fear I don't... The trouble I have is I don't know if I am allowed to evaluate specific factors inside the limit without having to replace all x's with the 0's, in my case. What I did in my post was simplifying by specifically evaluating the inverse tan, and then continuing, by using L'hopital's rule. Comparing to websites such as symbolab, they do not do this. They use L'hopital's rule without first simplifying arctan. – Freeze Aim Oct 27 '22 at 20:51
  • @Andrei: the rule works in an appropriate sense (the one required here) if just one of the limits is known to be finite and non-zero: if $g(x) \to a \in \Bbb{R}$ as $x \to x_0$ and $a \neq 0$, then $f(x)g(x)$ has a finite limit $b$, say, as $x \to x_0$ iff $f(x)$ has a finite limit, $c$ say. And, if so, we will have $b = ca$. – Rob Arthan Oct 27 '22 at 20:54
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    @Sambo On second thought, I do get it! What I did, inadvertedly, was split them into two different limits, x/sin3x and arctan (1/x), and then first evaluating arctan, and then evaluating the other limit, using L'hopital's rule. It makes sense! – Freeze Aim Oct 27 '22 at 20:58
  • What you did is correct. See [this answer](https://math.stackexchange.com/a/1783818/72031) for more details. – Paramanand Singh Oct 28 '22 at 01:37
  • @FreezeAim That's exactly right! Congrats on figuring it out yourself. If you want, you can post an answer to your own question. – Sambo Oct 28 '22 at 03:25

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