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

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?

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

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

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

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

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

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

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