# 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.

• Did you try with $x^{\cos x -1}$? – ecrin Jan 8 '19 at 16:19
• Hint: $(\ln{x})(\cos{x}-1) \rightarrow 0$. – Mindlack Jan 8 '19 at 16:20
• But the limit is still $0^0$ – violettagold Jan 8 '19 at 16:23

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 $$1$$.

\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}

$$\cos x= 1+o(1)$$ when $$x\to 0$$ (first order of Taylor expansion). Then $$\frac{x^{\cos x}}x =\frac{x^{1+o(1)}}x=\frac{x\cdot x^{o(1)}}x=x^{o(1)}\longrightarrow_{x\to0}1$$

• This is quite wrong: $y=x+o(1)$ does not imply $x^y=x+o(1)$, and $\frac{o(1)}x$ does not necessarily converge to $0$. – Did Jan 8 '19 at 16:48
• I should have corrected it now – Lorenzo B. Jan 8 '19 at 16:56
• Yeah, except that now, you pushed everything in the last step $x^{o(1)}\longrightarrow_{x\to0}1$, which happens to be wrong in general. Sorry but this is not a guessing game... – Did Jan 8 '19 at 16:59

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, $$\lim\limits_{x \to 0^{+}}\frac{x^{\cos x}}{x} = e^0 = 1$$.