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

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

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?

• Just to confirm, is it supposed to be $(\sqrt[3]{1 - \sin x})^2$? Aug 7, 2022 at 6:40
• yes sir, it is supposed to be just that Aug 7, 2022 at 6:42
• In that case, the limit does not exist, so I don't know how you computed it with L.H. Aug 7, 2022 at 6:44
• im sorry sir, wrong question Aug 7, 2022 at 6:46
• kindly ignore this Aug 7, 2022 at 6:46

Just a bit easier:

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

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!

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.

• Thankyou sir. ㅤ Aug 7, 2022 at 11:10