limit on infinity

Question

I am trying to solve this question but I am getting a negative infinity which is wrong. $$\lim _{x\to \infty }\left(\ln(e^{2x}-1)-\frac{x^2-3}{x}\right)$$ $$=\lim _{x\to \infty }\left(\ln(e^{2x}-1)-\frac{x(x^{ }-\frac{3}{x})}{x}\right)$$ $$=\lim _{x\to \infty }\left(\ln(e^{2x}-1)-x\right)$$ $$=\lim \:_{x\to \:\infty \:}\left((e^{2x}-1)\frac{\ln(e^{2x}-1)}{e^{2x}-1}-x\right)$$ $$\lim \:_{x\to \:\infty \:}\:(-x)$$ $$=-\infty $$

Answer 1

Just split up the $e^{2x}$ and you get

$$\ln(e^{2x}-1) - x+\frac 3x \stackrel{e^{2x}-1=e^{2x}(1-e^{-2x})}{=} \ln(e^{2x}) + \ln(1-e^{-2x}) - x+\frac 3x$$ $$=2x-x+\frac 3x + \ln(1-e^{-2x})=x+\frac 3x + \ln(1-e^{-2x})\stackrel{x \to +\infty}{\rightarrow}+\infty$$