What is $\lim_{x \to +\infty} \frac{(x+1)^x}{x^x}$? What is $$\lim_{x \to +\infty} \dfrac{(x+1)^x}{x^x}$$ and why? I believe it is $1$ because it is equal to $$\lim_{x \to +\infty}\dfrac{x^x}{x^x}$$
Wolfram|Alpha tells a different tale...
I know that the solution is $e$ and why, but what is wrong with my original thinking?
 A: Try exploring $$\left(1+\frac 1x\right)^x$$ using the binomial theorem - take $x$ as an integer $n$ to get $$1+n\cdot \frac 1n+\frac {n(n-1)}2\cdot \frac 1{n^2}+\dots=1+1+\frac 12\cdot\left(1-\frac 1n\right)+\dots$$
You can see that the limit is greater than $2$ (if it exists), and with a bit of work you will see that it increases to $e$.
Your original thinking is flawed, possibly because you have noticed that if you keep the exponent the same $$\cfrac {(1+x)^n}{x^n}=\left(1+\cfrac 1x\right)^n$$ the limit is $1$, because you can make the bit inside the bracket as close to $1$ as you like. But that doesn't work unless the exponent is under control.
A: Is $101^{100} \approx 100^{100}$? Can you do the approximation for larger $x$? 
I think Ian's hint is more than enough for you to get the answer.
A: Set $y=\frac{1}{x}$. Then, using l'Hôpital in the second line,
\begin{eqnarray*}
\frac{(x+1)^{x}}{x^{x}} &=&(1+\frac{1}{x})^{x}=\exp [\frac{1}{y}\ln (1+y)] \\
\lim_{y\rightarrow 0}\frac{\ln (1+y)}{y} &=&\lim_{y\rightarrow 0}\frac{1}{1+y%
}=1,
\end{eqnarray*}
so the limit equals $\exp [1]=e$.
