Without using L'Hopitals rule, how do you get $\lim_{x \to 0} \frac{1 - \cos 5x}{\sin 3x}$?

Question

Without using L'Hopitals rule, how do you get the limit

$$\lim_{x \to 0} \frac{1 - \cos 5x}{\sin 3x}$$

Using L'Hopital's, I can just take the derivative of the numerator $5 \sin 5x$ and denominator $3 \cos 3x$ which goes to $0/1$ or zero. But how do you show this without using derivatives?

Is there a trigonometric trick so that the denominator $\sin 3x$ is transformed so the term does not become $0/0$?

Any lead would be appreciated. Thanks

Answer 1

Note that $$ \frac{1-\cos 5x}{\sin 3x} = \frac{1-\cos 5x}{5x} \cdot \frac{5x}{3x} \cdot \frac{3x}{\sin 3x} $$ Now appeal to known special limits & limit properties.

Answer 2

@newbie py: Hi welcome to MSE. there are some methods to find this limit. like below $$\lim_{x\to 0}\frac{1-\cos(5x)}{\sin(3x)}\\=\\lim_{x\to 0}\frac{1-\cos(5x)}{\sin(3x)}\times\frac{1+\cos(5x)}{1+\cos(5x)}\\=\lim_{x\to 0}\frac{1-\cos^2(5x)}{\sin(3x)(1+\cos(5x))}\\\lim_{x\to 0}\frac{\sin^2(5x)}{\sin(3x)(1+\cos(5x))}$$ can you take over?

or use $1-\cos(u)=2\sin^2(\frac{u}{2})$ $$\lim_{x\to 0}\frac{1-\cos(5x)}{\sin(3x)}\\=\lim_{x\to 0}\frac{2\sin^2(\frac {5x}2)}{\sin(3x)}$$

Answer 3

We have that eventually

$$\cos 5x \ge \cos 6x=\frac{1-\tan^2 3x}{1+\tan^2 3x}$$

then

$$0\le \frac{1-\cos 5x}{|\sin 3x|}\le \frac{1-\cos 6x}{|\sin 3x|}=\frac{2|\sin 3x|}{(\cos ^23x)(1+\tan^2 3x)} \to \frac{0}{1\cdot (1+0)}=0$$