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How do I find the limit of

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

I know it is equal to $\frac{1}{4}$, but how did we get that? Without using L'Hopital's rule.
I tried canceling $x$ with $\sin(2x)$ and $\ln(x+1)$, but still got the wrong result...

19aksh
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Khalid Al Hajri
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  • Looks like I need a logarithms course... How do I expand it? – Khalid Al Hajri Feb 24 '21 at 06:07
  • https://math.stackexchange.com/questions/878374/taylor-series-of-ln1x – 19aksh Feb 24 '21 at 06:08
  • Hmm... I see lots of integrals and derivatives, but we haven't touched anything other than basics of derivatives... I got this question off of another student, so maybe we're just not supposed to solve this question right now? – Khalid Al Hajri Feb 24 '21 at 06:11
  • You will probably cover the proof afterwards, so for the time being just use the expansion formula, $\ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}$ – 19aksh Feb 24 '21 at 06:16
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    Your limit can be rearranged to $L_3$ in https://math.stackexchange.com/a/936642, which by demonstration is possible to solve without using L'Hopital or series expansion, but is quite complicated, and assumes the limits to exist beforehand. – player3236 Feb 24 '21 at 06:17
  • Yeah I just looked... If it's that complicated then probably I should just leave it for now. After all I am no MIT student... – Khalid Al Hajri Feb 24 '21 at 06:54

2 Answers2

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$$\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))} = \frac{0 }{ 0 },$$ you would make L'Hôpital's rule twice to have finally: $$\lim_{x \to 0} \frac{1/(x+1)}{ -4x \sin (2x) + 4 \cos (2x)} = 1/4.$$ Hope my answer helps you!

Gary
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Adib Akkari
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  • Why my answer is like this ???? Why latex is not working hhhhh loollllssssss ?! – Adib Akkari Feb 24 '21 at 07:17
  • You have to put your LaTeX code between two dollar signs (inline equation) or two double dollar signs (displayed equation). Note that the OP specifically told that the problem should be solved without using L'Hôpital's rule. – Gary Feb 24 '21 at 07:40
  • @Gary ok thx you – Adib Akkari Feb 24 '21 at 07:48
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here's my attempt to solve your problem without using L'Hospital rule :

Reminder :

$$\lim_{x\to0}\frac{\sin(ax)}{x}=a$$

From this post here you can see that : $$\lim_{x\to 0} \frac{\ln(x+1)-x}{x^2}=-\frac{1}{2}$$ We'll need these properties, so let's start :

Let our limit be $\Lambda$ \begin{align} \Lambda&=\lim_{x\to 0}\frac{x-\ln(x+1)}{x\sin(2x)}\\ &=\lim_{x\to 0} \frac{1}{\sin(2x)}\times \frac{x-\ln(x+1)}{x}\\ &=\lim_{x\to 0}\frac{x}{\sin(2x)}\times \frac{x-\ln(x+1)}{x^2}\\ &=\lim_{x\to 0}\frac{1}{\frac{\sin(2x)}{x}}\times \frac{x-\ln(x+1)}{x^2}\\ &=\frac{1}{2}\lim_{x\to 0} -\frac{\ln(x+1)-x}{x^2}\\ &=\frac{1}{2}\times \left((-1)\times \left(-\frac{1}{2}\right)\right)\\ &=\frac{1}{4} \end{align} Hence your limit $$\Lambda=\lim_{x\to 0}\frac{x-\ln(x+1)}{x\sin(2x)}=\frac{1}{4}$$

euler_med
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