I graphed this function on desmos and noticed that its value got close to $2.7$. However, the graph soon cut off because the numbers were too big for desmos to calculate. So, here is my attempt. We have $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$ and by Lhopital's rule, we have $\lim \limits_{x \to \infty} {(x+1)^x\ln(x+1)+1 \over x^x\ln(x)}$. I soon realized, using Wolfram that this won't work.
I start at the beginning with $\lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Let this limit be $L$. So, we have $L = \lim \limits_{x \to \infty} {(x+1)^x \over x^x}$. Taking natural log of both sides, we get $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$. I know that $x+1 \over x$ will get close to $1$ as $x$ gets very big, so $\ln({x+1 \over x})$ will approach $0$ when $x$ gets very big. Thus this limit $\ln(L) = \lim \limits_{x \to \infty} x\ln({(x+1) \over x})$ is a $0*\infty$ case, so I rewrite the limit as $\ln(L) = \lim \limits_{x \to \infty} {x \over \frac1{\ln({(x+1) \over x})}}$. I apply Lhopitals rule, but I end up with another complicated expression on the bottom using Wolfram, that is also an undefined limit. I have tried LHopital's rule twice and a "clever" manipulation, but the problem still won't yield, could someone please help? Thanks.
Best Answer
First note that $$\frac{(x+1)^x}{x^x}=\left(\frac{x+1}{x}\right)^x=\left(1+\frac{1}{x}\right)^x$$Even if you don't want to take $$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$$ as your definition of $e$, we can prove it is equal to $e$ by using L'Hopital's rule, as follows. I will use the notation $\exp(x)$ instead of $e^x$ to make the working more clear. By the way, I think you made a small error anyway: $\frac{d}{dx}x^x=x^x(\ln x+1)$, not $x^x\ln x$. $$\begin{align} \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x&=\lim_{x\to\infty}\exp\left(\ln\left(1+\frac{1}{x}\right)^x\right)\\ &=\lim_{x\to\infty}\exp\left(x\ln\left(1+\frac{1}{x}\right)\right)\\ &=\exp\left(\lim_{x\to\infty}x\ln\left(1+\frac{1}{x}\right)\right)\\ &=\exp\left(\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)\\ \end{align}$$ Now we are in a position to use L'Hopital's rule!
Hence we have $$\begin{align} \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x&=\exp\left(\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}}\right)\\ &=\exp\left(\lim_{x\to\infty}\frac{-\frac{1}{x^2}/(1+\frac{1}{x})}{-\frac{1}{x^2}}\right)\\ &=\exp\left(\lim_{x\to\infty}( 1+\frac{1}{x})\right)\\ &=\exp(1)=e \end{align} $$ as required! I hope that was helpful. If you have any questions, please don't hesitate to ask :)