[tex]\lim\limits_{x\to \infty} \Big(\ln(e^x+2^x)-\sqrt{x^2-4x+1}\Big) = \\ \\ = \lim\limits_{x\to \infty} \Big(\ln \big(e^x(1+\frac{2^x}{e^x})\big)-\sqrt{x^2-4x+1}\Big) = \\ \\ = \lim\limits_{x\to \infty} \Big(\ln e^x+\ln (1+\ffrac{2^x}{e^x})-\sqrt{x^2-4x+1}\Big) = \\ \\ = \lim\limits_{x\to \infty}\Big(x\ln e+\ln (1+\frac{2^x}{e^x})-\sqrt{x^2-4x+1}\Big)=[/tex]
[tex]= \lim\limits_{x\to 0}\Big(x-\sqrt{x^2-4x+1}\Big) +\lim\limits_{x\to \infty} \Big(\ln(1+\frac{2^x}{e^x})\Big)\\ \\ \lim\limits_{x\to \infty}{\frac{2^x}{e^x}} = \lim\limits_{x\to \infty}{{(\frac{2}{e})}^x}=0,\quad \frac{2}{e} < 1\ \\ =\lim\limits_{x\to\infty}\Big(x-\sqrt{x^2-4x+1}\Big) +0 = \\ \\ =\lim\limits_{x\to\infty}\Big(x-\sqrt{x^2-4x+1}\Big)= \\ \\ = \lim\limits_{x\to\infty} \dfrac{x^2-(x^2-4x+1)}{x+\sqrt{x^2-4x+1}}=\\ \\[/tex]
[tex]= \lim\limits_{x\to \infty} \dfrac{x(4-\frac{1}{x})}{x\Big(1+1\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}\Big)} = \dfrac{4}{2} = \boxed{2}[/tex]
