Question

lim((ln((1+x)/(1-x))-2x)/(x-senx))'x->0

Answer 1

$L=\lim_{x\to0}\dfrac{\ln\left(\dfrac{1+x}{1-x}\right)-2x}{x-\sin(x)}\\\\ \text{Por L'H\^ospital}:\\\\ L=\lim_{x\to0}\dfrac{\dfrac{1-x}{1+x}\cdot\dfrac{(1-x)-(-1)(1+x)}{(1-x)^2}-2}{1-\cos(x)}\\\\ L=\lim_{x\to0}\dfrac{\dfrac{1}{1+x}\cdot\dfrac{(1-x)+(1+x)}{(1-x)}-2}{1-\cos(x)}$

$L=\lim_{x\to0}\dfrac{\dfrac{2}{(1-x^2)}-2}{1-\cos(x)}\\\\ L=\lim_{x\to0}\dfrac{2-2(1-x^2)}{(1-\cos(x))(1-x^2)}\\\\ L=\lim_{x\to0}\dfrac{2x^2}{(1-\cos(x))(1-x^2)}$

$\text{Por L'H\^ospital de nuevo:}\\\\ L=\lim_{x\to0}\dfrac{4x}{\sin(x)(1-x^2)-2x(1-\cos(x))}\\\\ L=\lim_{x\to0}\dfrac{4x}{\sin(x)-x^2\sin(x)-2x+2x\cos(x)}\\\\ \text{Una vez m\'as:}\\\\ L=\lim_{x\to0}\dfrac{4}{\cos(x)-2x\sin(x)-x^2\cos(x)-2+2\cos(x)-2x\sin(x)}\\\\ L=\dfrac{4}{\cos(0)-2\cdot0\sin(0)-0^2\cos(0)-2+2\cos(0)-2\cdot0\sin(0)}\\\\ L=\dfrac{4}{1-0-0-2+2\cdot1-0}\\\\ L=\dfrac{4}{1-2+2}=\dfrac{4}{1}$

$\boxed{\lim_{x\to0}\dfrac{\ln\left(\dfrac{1+x}{1-x}\right)-2x}{x-\sin(x)}=4}$