Limit $\lim\limits_{x \to 0^{+}}\frac{x^{\cos x}}{x}$ I need to calculate the limit $\lim\limits_{x \to 0^{+}}\frac{x^{\cos x}}{x}$.
I tried to form it as $\lim\limits_{x \to 0^{+}}\frac{e^{\ln (x)\cdot \cos x}}{x} $ and do L'Hôpital's rule but it doesn't solve it.
 A: We have :
$$\frac{x^{\cos x}}{x} = x^{\cos x -1} = e^{\ln x (\cos x -1)}$$
Now using the fact that in a neighborhood of $0$ we have : 
$$\cos x - 1 = -\frac{x^2}{2} + o(x^2)$$
Then we can easily deduce that : 
$$\ln x \cdot (\cos x -1) \to 0$$
Hence the desired limit is $1$.
A: $$\begin{align}
\lim_{x\to0^+}x^{\cos x-1}&=\lim_{x\to0^+}e^{\ln x(\cos x-1)}\\
&=\exp\left(\lim_{x\to0^+}\frac{\cos x-1}{\dfrac1{\ln x}}\right)\\
&=\exp\left(\lim_{x\to0^+}\frac{-\sin x}{-\dfrac1{x(\ln x)^2}}\right)\\
&=\exp\left(\lim_{x\to0^+}(x\ln x)(\sin x\ln x)\right)=e^0
\end{align}$$
A: $\cos x= 1+o(1)$ when $x\to 0$ (first order of Taylor expansion). Then
$$\frac{x^{\cos x}}x =\frac{x^{1+o(1)}}x=\frac{x\cdot x^{o(1)}}x=x^{o(1)}\longrightarrow_{x\to0}1$$
A: You may also use the following facts:


*

*$\lim_{x\to 0}\frac{\cos x-1}{x} = \cos'(0) = -\sin(0) = 0$

*$\lim_{x\to 0}x\ln x = 0$
So, you get 
$$\ln \frac{x^{\cos x}}{x} = (\cos x - 1)\cdot \ln x = \frac{(\cos x - 1)}{x}\cdot x \ln x \stackrel{x\to 0^+}{\longrightarrow}0\cdot 0 = 0$$
Hence, $\lim\limits_{x \to 0^{+}}\frac{x^{\cos x}}{x} = e^0 = 1$.
