微积分咋用洛必达法则求极限(((x+1)^(1/x)*(1+1/x)^x)-4)/(x-1)^2呢？ 第1页

公式经过人工排版

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原公式

egin{align}& ext{该回答由算法自动生成}\\A&=limlimits_{x o1}frac{{left(frac{1}{x}+1 ight)}^x,{left(x+1 ight)}^{1/x}-4}{{left(x-1 ight)}^2}\&=limlimits_{x o1}frac{left(lnleft(frac{1}{x}+1 ight),{left(frac{1}{x}+1 ight)}^x-frac{{left(frac{1}{x}+1 ight)}^{x-1}}{x} ight),{left(x+1 ight)}^{1/x}+{left(frac{1}{x}+1 ight)}^x,left(frac{{left(x+1 ight)}^{frac{1}{x}-1}}{x}-frac{lnleft(x+1 ight),{left(x+1 ight)}^{1/x}}{x^2} ight)}{2,x-2}\&=limlimits_{x o1}frac{2,left(lnleft(frac{1}{x}+1 ight),{left(frac{1}{x}+1 ight)}^x-frac{{left(frac{1}{x}+1 ight)}^{x-1}}{x} ight),left(frac{{left(x+1 ight)}^{frac{1}{x}-1}}{x}-frac{lnleft(x+1 ight),{left(x+1 ight)}^{1/x}}{x^2} ight)-{left(frac{1}{x}+1 ight)}^x,left(frac{{left(x+1 ight)}^{frac{1}{x}-1}}{x^2}-frac{left(frac{1}{x}-1 ight),{left(x+1 ight)}^{frac{1}{x}-2}-frac{lnleft(x+1 ight),{left(x+1 ight)}^{frac{1}{x}-1}}{x^2}}{x}+frac{lnleft(x+1 ight),left(frac{{left(x+1 ight)}^{frac{1}{x}-1}}{x}-frac{lnleft(x+1 ight),{left(x+1 ight)}^{1/x}}{x^2} ight)}{x^2}+frac{{left(x+1 ight)}^{1/x}}{x^2,left(x+1 ight)}-frac{2,lnleft(x+1 ight),{left(x+1 ight)}^{1/x}}{x^3} ight)+{left(x+1 ight)}^{1/x},left(lnleft(frac{1}{x}+1 ight),left(lnleft(frac{1}{x}+1 ight),{left(frac{1}{x}+1 ight)}^x-frac{{left(frac{1}{x}+1 ight)}^{x-1}}{x} ight)-frac{lnleft(frac{1}{x}+1 ight),{left(frac{1}{x}+1 ight)}^{x-1}-frac{{left(frac{1}{x}+1 ight)}^{x-2},left(x-1 ight)}{x^2}}{x}+frac{{left(frac{1}{x}+1 ight)}^{x-1}}{x^2}-frac{{left(frac{1}{x}+1 ight)}^x}{x^2,left(frac{1}{x}+1 ight)} ight)}{2}\&=4,lnleft(2 ight)-3end{align}