I'm struggling a bit with solving a limit problem using L'Hopital's Rule:
$$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^{2x}$$
My work:
$$y = \left(1+\frac{1}{x}\right)^{2x}$$
$$\ln y = \ln \left(1+\frac{1}{x}\right)^{2x} = 2x \ln \left(1+\frac{1}{x}\right)$$
$$=\frac{\ln \left(1+\frac{1}{x}\right)}{(2x)^{-1}}$$
Taking derivatives of both the numerator and the denominator:
$$f(x) = \ln\left(1+\frac{1}{x}\right)$$
$$f'(x) = \left(\frac{1}{1+\frac{1}{x}}\right)\left(-\frac{1}{x^2}\right) = (x+1)\left(-\frac{1}{x^2}\right) = -\frac{(x+1)}{x^2}$$
$$g(x) = (2x)^{-1}$$
$$g'(x) = (-1)(2x)^{-2}(2) = -\frac{2}{(2x)^{2}} = -\frac{1}{2x^{2}}$$
Implementing the derivatives:
$$\lim_{x\to\infty} \frac{-\frac{(x+1)}{x^2}}{-\frac{1}{2x^2}} = -\frac{(x+1)}{x^2} \cdot \left(-\frac{2x^2}{1}\right) = 2(x+1) = 2x+2$$
However, I'm not sure where to go from here. If I evaluate the limit, it still comes out to infinity plus 2, and I don't know how much further to take the derivative or apply L'Hopital's Rule.
Any suggestions would be appreciated!
Best Answer
$$2\lim_{x\to\infty}\frac{\ln(1+x)-\ln x}{(x)^{-1}}$$
$$=2\lim_{x\to\infty}\frac{1/(1+x)-1/x}{-1/x^2}$$
$$=2\lim_{x\to\infty}\frac{x^2}{x(1+x)}$$
$$=2\lim_{x\to\infty}\frac1{1/x+1}=\frac2{0+1}$$