Find $\lim_{x\to0^+}\left(\frac{\ln(1+x)}{x}\right)^\frac{1}{x}$ 
Find $$\lim_{x\to0^+}\left(\frac{\ln(1+x)}{x}\right)^\frac{1}{x}.$$ 

I used some online calculators and found that the result is $\frac{1}{\sqrt e}$,but I can't find the way to solve this.
 A: Compute the limit of the logarithm, that is
$$
\lim_{x\to 0^+}\frac{\ln\dfrac{\ln(1+x)}{x}}{x}
$$
Since $\ln(1+x)=x-x^2/2+o(x^2)$, we get
$$
\lim_{x\to0^+}\frac{\ln(1-x/2+o(x))}{x}=
\lim_{x\to0^+}\frac{-x/2+o(x)}{x}=-\frac{1}{2}
$$
So the original limit is $e^{-1/2}$.
You can also use l’Hôpital:
\begin{align}
\lim_{x\to 0^+}\frac{\ln\dfrac{\ln(1+x)}{x}}{x}
&=\lim_{x\to 0^+}\frac{\ln\ln(1+x)-\ln x}{x} \\[6px]
&=\lim_{x\to 0^+}\left(\frac{1/(1+x)}{\ln(1+x)}-\frac{1}{x}\right)&&\text{[l’Hôpital]}\\[6px]
&=\lim_{x\to 0^+}\frac{x-(1+x)\ln(1+x)}{x(1+x)\ln(1+x)} \\[6px]
&=\lim_{x\to 0^+}\frac{x-(1+x)\ln(1+x)}{x\ln(1+x)} &&\text{[remove $1+x$]}\\[6px]
&=\lim_{x\to 0^+}\left(\frac{x-\ln(1+x)}{x\ln(1+x)}-1\right)\\[6px]
&=-1+\lim_{x\to 0^+}\frac{1-\dfrac{1}{1+x}}{\ln(1+x)+\dfrac{x}{1+x}}&&\text{[l’Hôpital]}\\[6px]
&=-1+\lim_{x\to0^+}\frac{x}{1+x}\frac{1+x}{(1+x)\ln(1+x)+x}\\[6px]
&=-1+\lim_{x\to0^+}\frac{x}{(1+x)\ln(1+x)+x}\\[6px]
&=-1+\lim_{x\to0^+}\frac{1}{\ln(1+x)+1+1}&&\text{[l’Hôpital]}\\[6px]
&=-1+\frac{1}{2}=-\frac{1}{2}
\end{align}
Which one is easier? ;-)
A: Let's try to use elementary techniques as far as possible. If the desired limit is $L$ then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0^{+}}\left(\frac{\log(1 + x)}{x}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0^{+}}\log\left(\frac{\log(1 + x)}{x}\right)^{1/x}\text{ (via continuity of log)}\notag\\
&= \lim_{x \to 0^{+}}\frac{1}{x}\log\left(\frac{\log(1 + x)}{x}\right)\notag\\
&= \lim_{x \to 0^{+}}\frac{1}{x}\left(\dfrac{\log(1 + x)}{x} - 1\right)\dfrac{\log\left(1 + \dfrac{\log(1 + x)}{x} - 1\right)}{\dfrac{\log(1 + x)}{x} - 1}\notag\\
&= \lim_{x \to 0^{+}}\frac{\log(1 + x) - x}{x^{2}}\cdot \lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (by putting }t = \frac{\log(1 + x)}{x} - 1)\notag\\
&= \lim_{x \to 0^{+}}\frac{\log(1 + x) - x}{x^{2}}\notag\\
&= \lim_{x \to 0^{+}}\dfrac{\dfrac{1}{1 + x} - 1}{2x}\text{ (via L'Hospital's Rule)}\notag\\
&= -\frac{1}{2}\lim_{x \to 0^{+}}\frac{1}{1 + x}\notag\\
&= -\frac{1}{2}\notag
\end{align}
Hence $L = e^{-1/2} = 1/\sqrt{e}$. Only one application of L'Hospital's Rule is needed.
A: Hint:
Write as
$$\exp\left(\lim\limits_{x\to0^{+}}\frac{1}{x}\ln\left(\frac{\ln(1+x)}{x}\right)\right)
\overset{\text{L'Hospital
}}{=}\exp\left(\lim\limits_{x\to0^{+}}\frac{x-\ln(x+1)-x\ln(x+1)}{x\ln(x+1)(x+1)}\right)$$
$$\exp\left(\overset{\text{L'Hospital}}{=} \lim\limits_{x\to 0^{+}}\frac{x}{x+\ln(x+1)+x\ln(x+1)}-1\right)$$
$$\exp\left(\overset{\text{L'Hospital}}{=} \lim\limits_{x\to 0^{+}}\frac{1}{1+\frac{1}{x+1}+\frac{x}{x+1}+\ln(x+1)}-1
\right)=\dots=\frac{1}{\sqrt e}$$
A: You have without problem $$\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$$ 
Put $h(x)=\frac{\ln (1+x)}{x}-1$ so that you get $$ (1+h(x))^{\frac 1x}=\left((        ((1+h(x))^{\frac 1x})^{\frac {1}{h(x)}}\right)^{\frac{h(x)}{x}}$$
    No problem now to find the limit $-\frac 12$ of $\frac {h(x)}{x}$, an exponent of the base clearly equal to $e$ (the easiest way by L’Hôpital).
