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

Question

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.

Answer 1

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 $e^L = e^0 = 1$

Answer 2

Hint:

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

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

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

Answer 3

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

Answer 4

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

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

Answer 5

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

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.