How to find the monotonicity of $\left ( 1+\frac{1}{x} \right ) ^{x}$ without derivation?

Question

The problem I ask for help is in the exercises of chapter 2, "limit" of my calculus book. When I tried to solve it, I found I have to prove the monotonicity of the function involved, shown in the question header. I cannot use techniques based on the derivative since this is the content of the following chapter so, at least in principle, the monotonicity of the function could be found without derivation, and for the same reason, many formulas get restricted such as inequalities proven through derivative.

Answer 1

Well it doesn't increase monotonically - check $ x \le 0$

So you want to obtain that $x_1 < x_2 => f(x_1) < f(x_2)$ for any $x_1, x_2 > 0$

Another stipulation - I'm not sure if you can prove this for the reals w/o differentiation, but it's certainly possible for the rationals, which can then also be extended to the reals. Thus, You need only show that $x_1 < x_2 => f(x_1) < f(x_2)$ where $x_2 = x_1 + 1$

Hence, check every term:

$x_1 < x_2 \implies (x_1)^{-1} > (x_2)^{-1}$

$\implies 1 + (x_1)^{-1} > 1 + (x_2)^{-1}$

So your $(1+1/x)$ is monotonic but decreasing.

Now you just need to figure out why exponentiating it to the x-power dominates your 1/x.

assume $x < x + 1$

Then just apply the formula:

$[(n+1)a-nb]b^n < a^{n+1}$

where $a = (1+\frac{1}{n+1})$ & $b = (1+\frac{1}{n})$

And it should check out I think.