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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?

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3 Answers 3

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

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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.

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


Alternatively,

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 $t\to1$

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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

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