# limit of $\cos{x}^{\frac{1}{\sin{x}}}$

I need to solve this limit:

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}}$$

I feel that it should be a rather easy limit, but I find myself struggling with the answer. I have tried substituting $$\cos{x}$$ and $$\sin{x}$$ with their series expansion, $$\cos{x} \approx 1-\frac{x^2}{2}+\dots$$ and $$\sin{x} \approx x-\frac{x^3}{6}+\dots$$, but the indeterminate form still remains.

• Can you clarify whether you mean $(\cos x)^{1/\sin x}$ or $\cos (x^{1/\sin x})$? Dec 4, 2020 at 13:47
• @MatthewvanEerde, the exercise is indeed a bit ambiguous, but by reading it I have assumed it is $(\cos{x})^(\frac{1}{\sin{x}})$. Dec 4, 2020 at 14:34

1. Call this function $$f(x)$$
2. Evaluate the limit $$L = \lim_{x \to 0} \log f(x) = \lim_{x \to 0}\log \left( \left(\cos x\right)^{1/ \sin x} \right) = \lim_{x \to 0} \left(\log \cos x\right) / \sin x$$ - this is a $$0/0$$ indeterminate form but you can apply L'Hôpital to get $$\lim_{x \to 0} \left(\left(1 / \cos x\right) \sin x\right) / \cos x = \lim_{x \to 0} \sin x / \cos^2 x = 0$$
3. The answer is $$e^L = e^0 = 1$$

Hint:

$$\lim_{x\to0}(\cos x)^{1/\sin x}$$

$$=\lim_{x\to0}(1-\sin^2x)^{1/2\sin x}$$

$$=\left(\lim_{x\to0}(1-\sin^2x)^{-\frac1{\sin^2x}}\right)^{-\lim_{x\to0}\frac{\sin x}2}$$

$$y=\big[\cos({x})\big]^{\frac{1}{\sin({x})}}\implies \log(y)={\frac{1}{\sin({x})}}\log(\cos(x))$$ Now, composing Taylor series $$1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(\cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$ $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ $$\log(y)=\frac{-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right) } {x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right) }=-\frac{x}{2}-\frac{x^3}{6}+O\left(x^5\right)$$ $$y=e^{\log(y)}=1-\frac{x}{2}+\frac{x^2}{8}+O\left(x^3\right)$$

• @Aryadeva. Sorry but $\log(cos(x))=-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}+O\left(x^8\right)$ Cheers :-) Dec 5, 2020 at 2:14
• Yes you're correct Claude. Got it. Thanks. +1 Dec 5, 2020 at 8:51

When $$\ (cos{x})^{\frac{1}{\sin{x}}}$$ is meant

$$\lim_{x\to 0}\, \cos{x}^{\frac{1}{\sin{x}}} = e^{\lim_{x\to 0}\,\frac{ln{cox{x}}}{\sin{x}}} = e^{\lim_{x\to 0}\,\frac{x*ln{cos{x}}}{\ x*sin{x}}} = e^{\lim_{x\to 0}\,\frac{ln{cos{x}}}{\ x}} = e^{\lim_{x\to 0}\,\frac{(cos{x} - 1) *ln{(cos{x} + 1 - 1)}}{\ (cos{x} - 1) * x}} = e^{\lim_{x\to 0}\,\frac{(cos{x} - 1)}{\ x}} = e^0 = 1$$

$$( \cos x)^{\frac{1}{ \sin x} }= ( 1 + ( \cos x -1) )^{\frac{1}{ \sin x} } = 1 + \frac{\cos x -1}{\sin x} + \frac{1}{\sin x} \left( \frac{1}{ \sin x} -1 \right) \frac{( \cos x -1)^2}{2} + \text{stuff}$$

Taking the limit as $$x \to 0$$ and using the fact that:

$$\lim_{ x \to 0} \frac{ (\cos x -1)^j}{ (\sin x)^k} =0$$

for $$j,k,n \in \mathbb{N}$$ and $$j \geq k$$

$$\lim_{x \to 0} ( \cos x)^{\frac{1}{ \sin x } } = 1$$

Explanation: The exponent is exploding off to infinity, but even for such a large exponent, we can use binomial series if the $$( \cos x -1) \leq 1$$ which it is for some interval around $$x=0$$. Use that and the second limit result I've written, you get the answer.

*:You can prove this by some maclaurain expansions.