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!

3 Answers

$$2\lim_{x\to\infty}\frac{\ln(1+x)-\ln x}{(x)^{-1}}$$

$$=2\lim_{x\to\infty}\frac{1/(1+x)-1/x}{-1/x^2}$$

$$=2\lim_{x\to\infty}\frac{x^2}{x(1+x)}$$

$$=2\lim_{x\to\infty}\frac1{1/x+1}=\frac2{0+1}$$

$\lim_{x \to \infty} (1+1/x)^{2x}$

$= \lim_{x \to \infty} \exp(\ln(1+1/x)^{2x})$

$= \lim_{x \to \infty} \exp(2x \ln(1+1/x))$

$= \lim_{x \to \infty} \exp(2 \frac{\ln(1+1/x)}{1/x})$

$= \exp(2 \lim_{x \to \infty} \frac{\ln(1+1/x)}{1/x}) \because e^{2x}$ is continuous.

Using LHR, we get

$= \exp(2 \lim_{x \to \infty} \frac{\frac{1}{(1+1/x)} (-1/x^2)}{-1/x^2})$

$=e^2$

Btw, I think you should know from PreCalculus (in a non-rigorous sense) that

$\lim_{x \to \infty} (1+1/x)^{x} = e$

Numerator= $\log\left(1+\dfrac1x\right)=\log(x+1)-\log(x)\to\dfrac1{x+1}-\dfrac1x=-\dfrac1{x(x+1)}$.

Denominator: $\dfrac1{2x}\to-\dfrac1{2x^2}$.

The ratio tends to $2$.