Matematické Fórum

stuart clark · 20. 07. 2011 20:47

$http://latex.codecogs.com/gif.latex?\hspace{-17}\lim_{x\rightarrow%20\frac{\pi}{2}}\frac{\cos%20x}{\left(1-\sin%20x\right)^{\frac{2}{3}}}$

vanok · 27. 10. 2011 12:55 — Editoval vanok (27. 10. 2011 18:19)

Hi ↑ stuart clark:,

I have just seen that this problem is not resolved
As I prefer to work on the neighborhood of 0, I put $X= x-\frac{\pi}2$ ( thus $x = \frac{\pi}2+X$ )
The expression
$\frac{\cos x}{(1-\sin x)^\frac23}$
becomes
$\frac{\cos $ \frac{\pi}2+X$}{$1-\sin ( \frac{\pi}2+X)$^\frac23}=\frac{- \sin(X)}{$1-\cos(X)$^\frac23}$
To find the limit asked I use equivalents in 0
For it I use

Autor skryl část textu. Zobrazit/skrýt!

$\sin(x) \sim_0 x$
$\cos(x) \sim_0 1-\frac{x^2}{2!}$

By using this we arrive towards
$\frac{- \sin(X)}{$1-\cos(X)$^\frac23}= \frac{-X}{$1- (1-\frac{X^2}{2!})$^{\frac23}}= \frac{-2^{2/3}}{X^{1/3}}$
We notice that the last expression have no limit in 0.
However limits to the left and to the right of 0 exist and are respectively $+\infty$ end $-\infty$ .
Obviously, it the case of your limit also

Sincerely Vanok

stuart clark · 27. 10. 2011 15:53

Thanks vanok you are saying Right.

Pavel · 27. 10. 2011 17:38 — Editoval Pavel (27. 10. 2011 17:39)

↑ vanok:

The limit can be determined in the more elementary way:

$\lim_{x\to\pi}\frac{\cos x}{(1-\sin x)^{\frac 23}}&=\lim_{x\to\pi}\biggl(\frac{\cos x}{(1-\sin x)^{\frac 23}}\cdot\frac{(1+\sin x)^{\frac 23}}{(1+\sin x)^{\frac 23}}\biggr)=2^{\frac 23}\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{(1-\sin^2x)^{\frac 23}} =2^{\frac 23}\lim_{x\to\frac{\pi}{2}}\frac{\cos x}{(\cos x)^{\frac 43}}\\ &=2^{\frac 23}\lim_{x\to\frac{\pi}{2}}\frac{1}{\sqrt[3]{\cos x}}\,.$

Since

$\lim_{x\to\frac{\pi}{2}^+}\frac{1}{\sqrt[3]{\cos x}}=-\infty\qquad\text{and}\qquad\lim_{x\to\frac{\pi}{2}^-}\frac{1}{\sqrt[3]{\cos x}}=\infty,$

the limit does not exist.