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Did I evaluate this limit correctly? My textbook did it like this, I'm guessing they just skipped the step to flip the trig functions?

$\lim_{x \to 0^+} \frac{x^2\csc3x}{1-\cos2x} $

$\lim_{x \to 0^+} \frac{x^2}{\sin(3x) (1-\cos2x)} $

$\frac{1}{3\cdot2} \lim_{x \to 0^+} \frac{3x}{\sin(3x)} \frac{2x}{(1-\cos2x)} $

$\frac{1}{3\cdot2} \lim_{x \to 0^+} \frac{1}{\frac{\sin(3x)}{3x}} \frac{1}{\frac{1-\cos2x}{2x}} $

$\frac{1}{6} \cdot 1 \cdot \infty = \infty$

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  • $\begingroup$ Both ways are correct. (Pretty much the same thing.) $\endgroup$ – KM101 Oct 29 '18 at 3:31
  • $\begingroup$ Is there some sort of short-cut or technique that the textbook applied, other than just skipping the flipping step? $\endgroup$ – Danielle Oct 29 '18 at 3:37
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The only difference in steps was this.

$$\frac{1}{\color{blue}{2}\cdot\color\purple{3}}\lim_{x \to 0^+} \frac{\color\purple{3}x}{\sin (3x)}\cdot \frac{\color\red{2}x}{1-\cos (2x)}$$

The book didn’t multiply the first limit by $3$ and multiply the entire expression by $\frac{1}{3}$ like you did. It just simplified. Both ways are correct.

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