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! :)

  • 1
    $\begingroup$ You forgot to multiply $4x$ by $3x$. $\endgroup$ – MathGod Dec 13 '14 at 17:57
  • $\begingroup$ Are you allowed using L'hopital? $\endgroup$ – user114138 Dec 13 '14 at 17:57
  • $\begingroup$ No, I'm not..., @MathGod, can you please point in which line? $\endgroup$ – 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.

  • $\begingroup$ Why you and I got different results? where was I wrong? $\endgroup$ – FigureItOut Dec 13 '14 at 18:07
  • $\begingroup$ I don't know; you have too many algebraic manipulations, which usually leads to a higher chance of making some mistake $\endgroup$ – 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: $$\frac{-1}{5(4/5-1)}=\frac{-1}{5(-1/5)}=1$$

  • $\begingroup$ @user1326293 edited $\endgroup$ – RE60K Dec 13 '14 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.