# Limit $\frac{0}{0}$ which tends to $\frac{\pi}{2}$

I'm trying to evaluate the following limit: $$\lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}$$ The limit has the form $\frac{0}{0}$, I've tried using L'Hopital's rule but I can't resolve it. Any idea?

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For a simple approach, we can use the following identities:

$1 - \sin 2x = (\cos x - \sin x)^2$

$\cos 2x = \cos ^2 x - \sin ^ 2 x$

Be careful when you consider the limits on either side.

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As $x\to\pi/2$, $\cos(x) \sim (\pi/2 - x)$. We also have $1 - \sin(x) = 1 - \cos(\pi/2 - x) \sim (1/2)(x - \pi/2)^2.$ Hence, $${\cos(x)\over (1 - \sin(x))^{2/3}}\sim {\pi/2 - x\over(1/2)^{2/3} (x - \pi/2)^{4/3}},$$ which blows up as $x\to\pi/2$.

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Multiply top and bottom by $(1+\sin x)^{2/3}$: $$\frac{\cos x}{(1-\sin x)^{2/3}}= \frac{\cos x(1+\sin x)^{2/3}}{(1-\sin x)^{2/3}(1+\sin x)^{2/3}}=\frac{\cos x(1+\sin x)^{2/3}}{(1-\sin^2 x)^{2/3}}=\frac{\cos x(1+\sin x)^{2/3}}{\cos^{4/3}x}.$$ Do the obvious cancellation. So we are interested in the behaviour of $$\frac{(1+\sin x)^{2/3}}{\cos^{1/3} x}$$ near $\pi/2$. For $x$ close to $\pi/2$, but less than $\pi/2$, this is large positive, while on the other side of $\pi/2$ it is large negative. Thus our limit does not exist.

Methodological comment: Taylor expansion, probably preceded by the change of variable $w=\pi/2-x$, remains the mathematically more natural approach. We are interested in local behaviour, and for nice functions that is precisely what the Taylor series gives us.

Pedantic comment: It is probably best to avoid non-integer powers of negative numbers. For one thing, the usual formal definition in terms of the logarithm breaks down. And when we deal with powers of negative numbers, familiar "rules" need to be used with caution.

So instead of $\frac{\cos x(1+\sin x)^{2/3}}{\cos^{4/3}x}$, it would have been better to write $\frac{\cos x(1+\sin x)^{2/3}}{|\cos x|^{4/3}}$. Then we can observe that this simplifies to $\frac{(1+\sin x)^{2/3}}{|\cos x|^{1/3}}$ if $\cos x$ is positive and to $\frac{-(1+\sin x)^{2/3}}{|\cos x|^{1/3}}$ if $\cos x$ is negative.

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If only one-sided limits can be calculated does it mean that limit does not exist ? – pedja Jan 16 '12 at 8:19
@pedja: If the one-sided limits are different, then the limit does not exist. But one could write $\lim_{x\to (\pi/2)^+} f(x)=-\infty$, $\lim_{x\to (\pi/2)^-} f(x)=+\infty$. – André Nicolas Jan 16 '12 at 15:22
,Since limit doesn't exist we can only say that function has non-zero oscillation at point $\pi/2$ – pedja Jan 16 '12 at 18:52
@pedja: We can write down our conclusion in a couple of ways. One is to say that the limit does not exist, and why. Another is to show that the left limit, right limit do exist (in the extended sense) but are different. – André Nicolas Jan 16 '12 at 19:22

$$\lim_{x\to \pi/2}\frac{\cos x}{(1-\sin x)^{2/3}}=\lim_{x\to \pi/2}\frac{\sin(\pi/2-x)}{(1-\cos(\pi/2-x))^{2/3}}=\lim_{t\to 0}\frac{\sin t}{(1-\cos t)^{2/3}}=\lim_{t\to 0}\frac{t+o(t)}{(t^2+o(t^2))^{2/3}}$$

The limit is $\infty$.

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if t is less zero. The limit will be $-\infty$. It means that limit doesnt exist. – Evgeny Savinov Jan 22 '12 at 22:20

$\displaystyle \lim_{x\rightarrow\pi/2}\frac{\cos(x)}{(1-\sin(x))^{2/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac{\cos^2{\frac{x}{2}}-\sin^2{\frac{x}{2}}}{(\sin^2{\frac{x}{2}}-2\sin{\frac{x}{2}}\cdot \cos {\frac{x}{2}}+\cos^2{\frac{x}{2}})^{2/3}}=$

$=\displaystyle \lim_{x\rightarrow\pi/2} \frac{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})(\cos {\frac{x}{2}}+\sin {\frac{x}{2}})}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{4/3}}=\displaystyle \lim_{x\rightarrow\pi/2} \frac {\cos {\frac{x}{2}}+\sin {\frac{x}{2}}}{(\cos {\frac{x}{2}}-\sin {\frac{x}{2}})^{1/3}}=\frac{\sqrt{2}}{0}=\infty$

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This is not quite right. Consider what happens when $x$ is close to $\pi/2$ but larger than $\pi/2$. – André Nicolas Jan 15 '12 at 21:37
There is no limit here. – Evgeny Savinov Jan 22 '12 at 22:21