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?
Asked
Active
Viewed 183 times
4 Answers
2
$$\frac{\cos x}{(1-\sin x)^\frac23}=\frac{\cos x(1+\sin x)^\frac23}{(1-\sin^2x)^\frac23}=\frac{(1+\sin x)^\frac23}{(\cos x)^\frac13}$$
Mike
- 13,318
-
I think you have made a mistake in your third step. $(1-\sin^2{x})^{2/3} = (\cos{x})^{4/3}$ – Ron Gordon Feb 03 '13 at 12:20
-
@rlgordonma Already fixed. I deleted it temporarily as it led to the wrong conclusion. – Mike Feb 03 '13 at 12:22
1
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.
Ron Gordon
- 138,521
-
-
No, it goes the opposite way, to $\infty$, which means it increases without bound. – Ron Gordon Feb 03 '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 03 '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 03 '13 at 20:40
1
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 $t\to1$
lab bhattacharjee
- 274,582
-
-
@vonbrand, ya, the limit does not converge to a finite value, hence is divergent. – lab bhattacharjee Feb 04 '13 at 03:27
0
To evaluate the limit:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
we can use the fact that as $x$ approaches $\frac{\pi}{2}$, $\sin x$ approaches 1 and $\cos x$ approaches 0. Substituting these values into the expression, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{\left(1-\sin x\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule to evaluate the limit. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-3\left(1-\sin x\right)^{-\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{3\left(-\frac{1}{3}\right)\left(1-\sin x\right)^{-\frac{4}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{-1\left(1-\sin x\right)^{\frac{4}{3}}}\right)$$
Again, the expression is indeterminate in this form ($\frac{0}{0}$), so we can try using L'Hopital's rule again. Differentiating the numerator and denominator with respect to $x$ gives us:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\sin\left(x\right)}{-4\left(1-\sin x\right)^{\frac{1}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{-\cos\left(x\right)}{4\left(\frac{1}{3}\right)\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
$$= \lim_{x\to \frac{\pi }{2}}\left(\frac{\cos\left(x\right)}{1\left(1-\sin x\right)^{\frac{2}{3}}}\right)$$
Substituting the values $\sin x = 1$ and $\cos x = 0$, we get:
$$\lim_{x\to \frac{\pi }{2}}\left(\frac{0}{\left(1-1\right)^{\frac{2}{3}}}\right) = \lim_{x\to \frac{\pi }{2}}\left(\frac{0}{0^{\frac{2}{3}}}\right)$$
The expression is indeterminate in this form ($\frac{0}{0}$), so we can't use L'Hopital's rule any further. However, since $\cos x$ approaches 0 and $\left(1 - \sin x\right)^{\frac{2}{3}}$ approaches 0 as $x$ approaches $\frac{\pi}{2}$, the entire expression approaches $\frac{0}{0}$, which is indeterminate. Therefore, the limit does not exist
mayyas haj
- 131