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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! :)

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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: $$\frac{-1}{5(4/5-1)}=\frac{-1}{5(-1/5)}=1$$

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

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