For every positive real number $x$, let
\[
g(x)=\lim_{r\to 0}  ((x+1)^{r+1}-x^{r+1})^{\frac{1}{r}}.
\]
Find $\lim_{x\to \infty}\frac{g(x)}{x}$.

[hide=Solution]
By the Binomial Theorem one obtains\\
$\lim_{x \to \infty} \lim_{r \to 0} \left((1+r)+\frac{(1+r)r}{2}\cdot x^{-1}+\frac{(1+r)r(r-1)}{6} \cdot x^{-2}+\dots \right)^{\frac{1}{r}}$\\
$=\lim_{r \to 0}(1+r)^{\frac{1}{r}}=\boxed{e}$
[/hide]
1. We start with the given function:
   \[
   g(x) = \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{\frac{1}{r}}
   \]

2. To simplify this, we take the natural logarithm of \( g(x) \):
   \[
   \log(g(x)) = \log \left( \lim_{r \to 0} \left( (x+1)^{r+1} - x^{r+1} \right)^{\frac{1}{r}} \right)
   \]
   Using the property of logarithms, we get:
   \[
   \log(g(x)) = \lim_{r \to 0} \frac{\log \left( (x+1)^{r+1} - x^{r+1} \right)}{r}
   \]

3. We apply L'Hopital's Rule to evaluate the limit:
   \[
   \log(g(x)) = \lim_{r \to 0} \frac{(x+1)^{r+1} - x^{r+1}}{r}
   \]
   Differentiating the numerator and the denominator with respect to \( r \):
   \[
   \log(g(x)) = \lim_{r \to 0} \frac{(x+1)^{r+1} \log(x+1) - x^{r+1} \log(x)}{1}
   \]
   Evaluating the limit as \( r \to 0 \):
   \[
   \log(g(x)) = (x+1) \log(x+1) - x \log(x)
   \]

4. Exponentiating both sides to solve for \( g(x) \):
   \[
   g(x) = e^{(x+1) \log(x+1) - x \log(x)}
   \]
   Simplifying the exponent:
   \[
   g(x) = \frac{(x+1)^{x+1}}{x^x}
   \]

5. We now need to find:
   \[
   \lim_{x \to \infty} \frac{g(x)}{x}
   \]
   Substituting \( g(x) \):
   \[
   \lim_{x \to \infty} \frac{(x+1)^{x+1}}{x^x \cdot x} = \lim_{x \to \infty} \frac{(x+1)^{x+1}}{x^{x+1}}
   \]

6. Simplifying the expression inside the limit:
   \[
   \lim_{x \to \infty} \left( \frac{x+1}{x} \right)^{x+1} = \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^{x+1}
   \]

7. Recognizing the limit as the definition of \( e \):
   \[
   \lim_{x \to \infty} \left( 1 + \frac{1}{x} \right)^{x+1} = e
   \]

The final answer is \( \boxed{e} \).