Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can you help me to solve this limit? $\frac{\cos x}{(1-\sin x)^{2/3}}$... as $x \rightarrow \pi/2$, how can I transform this?

share|cite|improve this question

$$\frac{\cos x}{(1-\sin x)^\frac23}=\frac{\cos x(1+\sin x)^\frac23}{(1-\sin^2x)^\frac23}=\frac{(1+\sin x)^\frac23}{(\cos x)^\frac13}$$

share|cite|improve this answer
I think you have made a mistake in your third step. $(1-\sin^2{x})^{2/3} = (\cos{x})^{4/3}$ – Ron Gordon Feb 3 '13 at 12:20
@rlgordonma Already fixed. I deleted it temporarily as it led to the wrong conclusion. – Mike Feb 3 '13 at 12:22

Hint: let $y = \pi/2 - x$ and take the limit as $y \rightarrow 0$.

In this case, the limit becomes

$$\lim_{y \rightarrow 0} \frac{\sin{y}}{(1-\cos{y})^{2/3}}$$

That this limit diverges to $\infty$ may be shown several ways. One way is to recognize that, in this limit, $\sin{y} \sim y$ and $1-\cos{y} \sim y^2/2$, and the limit becomes

$$\lim_{y \rightarrow 0} \frac{2^{2/3} y}{y^{4/3}} = \lim_{y \rightarrow 0} 2^{2/3} y^{-1/3} $$

which diverges.

share|cite|improve this answer
so the limit is zero? – Kyle92 Feb 3 '13 at 12:13
No, it goes the opposite way, to $\infty$, which means it increases without bound. – Ron Gordon Feb 3 '13 at 12:13
Use l'Hôpital? That's the first thing I'd try. Or expand by Taylor around the limiting value... – vonbrand Feb 3 '13 at 20:12
@vonbrand: I do not understand your point. I did precisely the latter of your options. I simply moved the limit point to the origin as it makes things easier to look at. – Ron Gordon Feb 3 '13 at 20:40

As $\cos x=\cos^2\frac x2-\sin^2\frac x2$ and $1-\sin x=(\cos\frac x2-\sin \frac x2)^2$

$$\lim_{x\to\frac\pi2}\frac{\cos x}{(1-\sin x)^\frac23}$$

$$=\lim_{x\to\frac\pi2}\frac{(\cos\frac x2-\sin\frac x2)(\cos\frac x2+\sin\frac x2)}{(\cos\frac x2-\sin \frac x2)^\frac43}$$

$$=\lim_{x\to\frac\pi2}\frac{(\cos\frac x2+\sin\frac x2)}{(\cos\frac x2-\sin \frac x2)^\frac13} \text{ which is of the form } \frac{\sqrt2}0$$

as $x\to \frac\pi2, \frac x2\to \frac\pi4\implies \tan \frac x2\to1 \implies \tan \frac x2\ne1\implies \cos \frac x2\ne \sin\frac x2$


putting $t=\tan\frac x2$ so that $x\to\frac\pi2,t\to1$ and $$\cos x=\frac{1-\tan^2\frac x2}{1+\tan^2\frac x2}=\frac{1-t^2}{1+t^2}\text {and } \sin x=\frac{2\tan\frac x2}{1+\tan^2\frac x2}=\frac{2t}{1+t^2},1-\sin x=\frac{(1-t)^2}{1+t^2}$$

So, $$\lim_{x\to\frac\pi2}\frac{\cos x}{(1-\sin x)^\frac23}$$

$$=\lim_{t\to1}\frac{(1-t^2)}{(1+t^2)}\cdot \frac{(1+t^2)^\frac23}{(1-t)^\frac43}$$

$$=\lim_{t\to1}\frac{(1+t)}{(1+t^2)^\frac13(1-t)^\frac13} \text{ which is of the form } \frac10$$

as $t\ne1$ as $t\to1$

share|cite|improve this answer
Then your limit is $\infty$... – vonbrand Feb 3 '13 at 22:53
@vonbrand, ya, the limit does not converge to a finite value, hence is divergent. – lab bhattacharjee Feb 4 '13 at 3:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.