# Evaluating trigonometric limits $\frac{2x-\sin(3x)}{4x-\sin(5x)}$

I need to evaluate this limit:

$$\lim_{x \to 0 }\frac{2x-\sin(3x)}{4x-\sin(5x)}$$

I did: $$\frac{2x}{4x-\sin(5x)}-\frac{\sin(3x)3x}{(4x-\sin(5x))3x}$$

$$\frac{2x}{4x-\sin(5x)}-\frac{3x}{4x-\sin(5x)}$$

$$\frac{-x}{4x-\sin(5x)}$$

$$\frac{-5x}{\sin(5x)(\frac{4x}{\sin(5x)}-1)*5}$$

$$\frac{-1}{(\frac{4x}{\sin(5x)}-1)5} =\frac{1}{5}$$

I know that I did wrong since the answer is 1.. What did I do wrong? How can you evaluate this limit?

Thanks! :)

• You forgot to multiply $4x$ by $3x$. – MathGod Dec 13 '14 at 17:57
• Are you allowed using L'hopital? – user114138 Dec 13 '14 at 17:57
• No, I'm not..., @MathGod, can you please point in which line? – FigureItOut Dec 13 '14 at 17:59

It's easier to write it as $\frac{2x(1-\frac{\sin 3x}{2x})}{4x(1-\frac{\sin 5x}{4x})}$, then do a bit of algebra with the $\frac{\sin x}{x}$ limit and get the result.

• Why you and I got different results? where was I wrong? – FigureItOut Dec 13 '14 at 18:07
• I don't know; you have too many algebraic manipulations, which usually leads to a higher chance of making some mistake – Alex Dec 13 '14 at 18:10

You did:

$$\frac{2x}{4x-\sin(5x)}-\frac{\sin(3x)3x}{(4x-\sin(5x))3x}\\ \frac{2x}{4x-\sin(5x)}-\frac{3x}{4x-\sin(5x)}\\ \frac{-x}{4x-\sin(5x)}\\ \frac{-5x}{\sin(5x)(\frac{4x}{\sin(5x)}-1)*5}\\ \frac{-1}{\color{red}{(\frac{4x}{\sin(5x)}}-1)5} =\frac{1}{5}\\$$

But: $$\lim_{x\to0}\frac{4x}{\sin(5x)}=\lim_{x\to0}\frac{5x}{\sin 5x}\frac{4x}{5x}=\frac45$$

So the limit is: $$\frac{-1}{5(4/5-1)}=\frac{-1}{5(-1/5)}=1$$

• @user1326293 edited – RE60K Dec 13 '14 at 18:13