Finding limit of $\left(1 +\frac1x\right)^x$ without using L'Hôpital's rule. Is there a way to find this limit without using L'Hôpital's rule. Just by using some basic limit properties.
$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e$$
 A: We may show that
$$ \lim_{x\to +\infty} x\left(\log(x+1)-\log(x)\right) = 1 \tag{1}$$
by noticing that
$$ x\left(\log(x+1)-\log(x)\right)=\int_{x}^{x+1}\frac{x}{t}\,dt \tag{2}$$
is trivially bounded between $\frac{x}{x+1}$ and $1$, since $f(t)=\frac{1}{t}$ is a decreasing function on $\mathbb{R}^+$.
A: In the comments it has been pointed out that this usually serves as a definition of e.  But I think I know what you're asking.  Usually in introductory calculus classes to evaluate this limit you let $$y= \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x,$$ and then take the natural log of both sides to get $$\ln(y) = \lim_{x\to\infty} x\ln\left(1+\frac{1}{x}\right)=\lim_{x\to\infty}\frac{\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x}},$$ where the last limit you use L'Hôpital's rule to get that it is equal to 1.  So I think maybe what you are actually asking is can, $$\lim_{x\to\infty} x\ln\left(1+\frac{1}{x}\right)$$ be evaluated without L'Hôpital's rule?  Please correct me if I'm wrong.  If this is indeed your question you can expand $\ln\left(1+\frac{1}{x}\right)$ as a series at $x=\infty$ which I think makes the limit trivial.  
I don't think I've ever used these kinds of power series so I would appreciate any input about this last suggestion.
A: Here are two of my questions
which, combined,
answer your question:
Is this proof that $\lim_{n \to \infty} (1+1/n)^n$ exists (1) new, (2) interesting?
Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists
