A question on Trigonometric limits

Question

What will this expression $(\cos x)^{\csc x}$ simplify to when $x\rightarrow 0$. I tried writing $\cos{x}$ as $(1-(1-\cos{x}))$

Answer 1

Note that $$ L= \log\left(\cos x\right)^{\csc x}= \csc x\log\cos x= \frac{\log\cos x}{\sin x} \to\frac{(\log\cos x)'}{(\sin x)'}= \frac{\frac{-\sin x}{\cos x}}{\cos x} \to0 $$ by L'Hopital's Rule, and therefore $$\left(\cos x\right)^{\csc x}=e^L\to e^0=1$$ as $x\to0$. If we're dead set against using L'Hopital's Rule, then we can expand some Taylor (or Maclaurin) series: $$\begin{align}L&=\log\left(\cos x\right)^{\csc x}\\ &=\csc x\cdot\log\cos x\\ &=\left(\frac1x+\tfrac16x+\tfrac7{360}x^3+\cdots\right) \left(-\tfrac12x^2-\tfrac1{12}x^4-\tfrac1{45}x^6+\cdots\right)\\ &=-\left(\tfrac12x+\tfrac16x^3+\tfrac{11}{240}x^5+\tfrac{11}{840}x^7+\cdots\right) \end{align} $$ which will also clearly approach $0$ as $x$ does, since the constant term is $0$.

Answer 2

$y = \lim_\limits{x\to0} (\cos x)^{\csc x}\\ \ln y = \lim_\limits{x\to0} \frac{\ln(\cos x)}{sin x}$

Now I generally avoid L'Hopital's rule. But, this would be an ideal time to use it.

$\ln y = \lim_\limits{x\to0} \frac{\tan x)}{cos x} = 0$ $y = 1$

However, you have tagged this "limits without L'Hopitals" I will do my best.

when $0.5<x<1.5, |\ln x| < |1-x^2|$

$|\ln(\cos x)| < 1-\cos^2 x = \sin^2 x$ when $|x|<|\pi/3|$

$|\ln y| < |\lim_\limits{x\to0} \sin x|$

And $\ln y$ gets squeezed to 0.

Answer 3

Bernoulli's Inequality guarantees $$ \begin{align} \lim_{x\to0}\cos(x)^{\csc(x)} &=\lim_{x\to0}\left(1-\sin^2(x)\right)^{\frac1{2\sin(x)}}\\ &\ge\lim_{x\to0}(1-\sin(x))^{1/2}\\[2pt] &=1 \end{align} $$