How do I find the limit of $\lim_{x \to 0} \frac{x-\ln(x+1)}{x(\sin(2x))}$?

Question

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...

Answer 1

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

Answer 2

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 $$\Lambda=\lim_{x\to 0}\frac{x-\ln(x+1)}{x\sin(2x)}=\frac{1}{4}$$