calculus - L'Hopital's Rule to determine limit as x approaches infinity

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!