I'm having big troubles trying to solve this limit:
$$\lim_{x\to 0} \quad x(2-x)\sqrt{\frac{1}{x^2}-\frac{4e^{-2x}}{(1-e^{-2*x})^{2}}} + x$$
The result is zero, but I can't figure out how to get there.
I was trying to make a "sandwich" by making an upper and lower approximation using the fact that $1/e < 1/(1+x)$ but this is getting me nowhere. Rationalisation isn't helping either.
Thank you in advance for your inputs
Bests
Note that
$$x(2-x)\sqrt{\frac{1}{x^2}-\frac{4e^{-2x}}{(1-e^{-2*x})^{2}}} + x=\frac{x(x-2)}{|x|}\sqrt{1-\frac{(2x)^2e^{-2x}}{(1-e^{-2x})^{2}}}+x=\frac x{|x|}(x-2)\sqrt{1-\left(\frac{2x}{1-e^{-2x}}\right)^2e^{-2x}}+x\to \pm1\cdot(-2)\cdot0+0=0$$
indeed $y=-2x\to 0$
$$\frac{2x}{1-e^{-2x}}=\frac{y}{e^{y}-1}\to 1$$