Evaluating $\lim _{x\rightarrow 0}\left( \cos x\right) ^{\cot ^{2}x}$

Question

This math has one trigonometric ratio as power of another trigonometric function.

$$\lim _{x\rightarrow 0}\left( \cos x\right) ^{\cot ^{2}x}$$

Answer 1

Considering $$y=\big(\cos(x)\big) ^{\cot ^2(x)}\implies \log(y)=\cot ^2(x)\log(\cos(x))$$ use composition of Taylor series $$\cos(x)=1-\frac{x^2}{2}+O\left(x^4\right)$$ $$\log(\cos(x))=-\frac{x^2}{2}+O\left(x^4\right)$$ $$\cot(x)=\frac{1}{x}-\frac{x}{3}+O\left(x^3\right)$$ $$\cot ^2(x)=\frac{1}{x^2}-\frac{2}{3}+O\left(x^2\right)$$ $$\log(y)=-\frac{1}{2}+O\left(x^2\right)$$ $$y=e^{\log(y)}=\frac{1}{\sqrt{e}}+O\left(x^2\right)$$

If you want to see more, add one more term to each expansion; this would give $$y=\frac{1}{\sqrt{e}}\left(1+\frac{x^2}{4}+O\left(x^4\right) \right)$$ which, for sure, shows the limit but also how it is approached.

Moreover, this is a quite good approximation. Using $x=\frac \pi 6$ (this is quite far away from $0$), the exact value is $y=\frac{3 \sqrt{3}}{8}\approx 0.649519$ while the last truncated expansion gives $\frac{144+\pi ^2}{144 \sqrt{e}}\approx 0.648102$ which corresponds to a relative error of $0.2$%.

Answer 2

Hint. Use that $$e^{\lim_{x\to 0}\frac{\ln(\cos(x))}{\tan^2(x)}}$$

Answer 3

Hint: $$\lim_{x\to0}(\cos x)^{\cot^2x}=\left(\left(\lim_{x\to0}(1+\cos x-1)^{1/(\cos x-1)}\right)^{\lim_{x\to0}(\cos x-1)/\sin^2x}\right)^{\lim_{x\to0}\cos^2x}$$

Now $\lim_{x\to0}\dfrac{\cos x-1}{\sin^2x}=-\lim_{x\to0}\dfrac{\cos x-1}{\cos^2x-1}=-\lim_{x\to0}\dfrac1{1+\cos x}=?$

Answer 4

You can also do it as follows using $\lim_{t\to 0}(1-t)^{\frac 1t}=\frac 1e$.

Write $\cos x = 1 -2\sin^2 \frac x2$, so you get

\begin{eqnarray*} \left( \cos x\right)^{\cot x} & = & \left(\underbrace{\left(1-2\sin^2\frac x2\right)^{\frac{1}{2\sin^2 \frac x2}}}_{\stackrel{x\to 0}{\rightarrow}\frac 1e}\right)^{\underbrace{\frac{2\sin^2 \frac x2}{\sin ^2 x}}_{\stackrel{x\to 0}{\rightarrow}\frac 12}\cdot \cos^2 x}\\ & \stackrel{x\to 0}{\rightarrow} & \frac{1}{\sqrt e} \end{eqnarray*}