Calculate the following limit: $\lim\limits_{x \to 0^+}\!\big((1+x)^x-1\big)^x $

Question

This limit is from a college admission exam in Cluj, Romania. I've tried writing the limit as $\,e^{g\cdot\ln f}$, so $\,e^{\lim\limits_{x \to 0^+}x\cdot\ln\left((1+x)^x-1\right)}$, but then I have no idea how to write $(1+x)^x$.

Answer 1

Using $\lim_{u\to 0} \frac{e^u-1}{u} = 1$ and $\lim_{x\to 0} x\ln(1+x) = 0$ along with the fact $x\ln(1+x) \neq 0$ whenever $x\neq 0$, we can write $$\lim_{x\to 0^+}((1+x)^x - 1)^x = \lim_{x\to 0^+}\left(\frac{e^{x\ln(1+x)}-1}{x\ln(1+x)} \cdot (x\ln(1+x))\right)^x = 1^0 \cdot \lim_{x\to 0^+}(x\ln(1+x))^x$$

Now, using $\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1$, we can write $$\lim_{x\to 0^+}(x\ln(1+x))^x = \lim_{x\to 0^+}\left(x^2 \cdot \frac{\ln(1+x)}{x}\right)^x = \lim_{x\to 0^+} (x^2)^x \cdot 1^0 = \lim_{x\to 0^+}x^{2x}$$

Showing $\lim_{x\to 0^+}x^{2x} = 1$ is routine.

Answer 2

Over positive $x$ $$((1+x)^x-1)^x=(e^{x\ln(x+1)}-1)^x=e^{x\ln((e^{x\ln(x+1)}-1))}$$

When $x$ goes to $0^+$. First $x\ln(x+1)$ goes to 0 then :

$$e^{x\ln(x+1)}-1=1+x\ln(x+1)-1+o_0(x\ln(x+1))$$

So

$$x\ln((e^{x\ln(x+1)}-1)) \to_{0^+} 0$$

Then

$$((1+x)^x-1)^x=(e^{x\ln(x+1)}-1)^x=e^{x\ln((e^{x\ln(x+1)}-1))} \to_{0^+} 1$$

Answer 3

Let's expand everything using the binomial theorem. As $x\to 0^+,$ $$((1+x)^x-1)^x\approx ((1+x^2)-1)^x=x^{2x}$$ so $$((1+x)^x-1)^x\to x^{2x}\to 1.$$ Note: $$(1+x)^x=1 + x^2 - \frac{x^3}{2} +O(x^4)$$