How to compute $\lim_{x\to{\pi/2}} \frac{\cos{x} }{\sqrt[3]{(1-\sin{x})^2}}$

Question

$$\lim_{x\to{\pi/2}} \frac{\cos{x} }{\sqrt[3]{(1-\sin{x})^2}}$$

I am on terms with high school level calculus, and can solve this using L-hospital's method, but cannot come up with any other method to do the same. Please can someone tell me about any alternative methods to solve this question?

Answer 1

Just a bit easier:

\begin{eqnarray*} \frac{\cos x}{\sqrt[3]{(1 - \sin{x})^2}} & = & \sqrt[3]{(1 + \sin{x})^2} \frac{\cos x}{\sqrt[3]{\cos^4{x}}}\\ & = & \frac{\sqrt[3]{(1 + \sin{x})^2}}{\sqrt[3]{\cos x}}\\ & \stackrel{x \to (\frac{\pi}{2})^-}{\longrightarrow} & +\infty \end{eqnarray*}

Answer 2

Use standard trigonometrical formulae and write the function of the limit (setting $y=tan(\dfrac{x}{2}$))

$\dfrac{1-y^{2}}{1+y^{2}}$.$\dfrac{1}{(1-\dfrac{2y}{1+y^{2}})^{2/3}}$=

=$\dfrac{(1-y^{2})(1+y^{2})^{2/3}}{(1+y^{2})(1-y)^{4/3}}$

$=(1+y^{2})^{-1/3}\dfrac{1+y}{(1-y)^{1/3}}=f(y)$.

Since $y\to 1$ we clearly see that $\displaystyle \lim_{y \to 1^{+}}f(y)=-\infty$ and $\displaystyle \lim_{y \to 1^{-}}f(y)=+\infty$

therefore the limit does not exist!

Answer 3

For $x \to (\frac{\pi}{2})^-$ we have $\cos x = (1 - \sin^2{x})^{1/2}$ and therefore: \begin{align*} \lim_{x \to (\frac{\pi}{2})^-} \frac{\cos x}{\sqrt[3]{(1 - \sin{x})^2}} &= \lim_{x \to (\frac{\pi}{2})^-} \frac{(1 - \sin^2{x})^{1/2}}{(1 - \sin{x})^{2/3}}\\ &= \lim_{x \to (\frac{\pi}{2})^-} \frac{(1 - \sin{x})^{1/2} \times (1 + \sin{x})^{1/2}}{(1 - \sin{x})^{2/3}}\\ &= \lim_{x \to (\frac{\pi}{2})^-} \frac{(1 + \sin{x})^{1/2}}{(1 - \sin{x})^{1/6}}\\ &= +\infty \end{align*} But if $x \to (\frac{\pi}{2})^+$, then $\cos{x} = -(1 - \sin^2{x})^{1/2}$ and the same calculation shows that \begin{align*} \lim_{x \to (\frac{\pi}{2})^+} \frac{\cos x}{\sqrt[3]{(1 - \sin{x})^2}} = -\infty \end{align*} So the given limit doesn't exist.