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I'm trying to use L'Hopital's rule to calculate:

$$\lim_{x \to 0^+} \dfrac{x - \sin x}{(x \sin x)^{(3/2)}}$$

Taking a couple of derivatives of the denominator gets quite nasty, so I'd like to find a simpler way to do it.

I would like to make a change of variable, say, $t = \sqrt{x \sin x}$, to get a $t^3$ in the denominator. Unfortunately, that leaves me with problems in the numerator. Maybe there is some other manipulation or some trig identity that simplifies things that I am missing? This shouldn't be a difficult problem, but I can't seem to find a slick way to do it.

The answer is given as $\frac{1}{6}$. Thanks for your help.

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I added a 2nd answer alongside my first. The new answer is better. –  alex.jordan Apr 13 '12 at 2:27
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3 Answers

up vote 3 down vote accepted

Instead of taking multiple derivatives, maybe stop after the taking one derivative. Then think about what to do with the $1-\cos x$ in the numerator. Perhaps you know what $$ \lim_{x\rightarrow 0}\frac{1-\cos x}{x^2} $$ is? If not, try multiplying numerator and denominator by $1+\cos x$.

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I'm not sure I see where this gets me. I take the first derivative and have $\dfrac{1 - \cos x}{\frac{3}{2} \sqrt{x \sin x} (x \cos x + \sin x)}.$ I can certainly get $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$. I'm sure I'm missing something here. –  user23784 Apr 13 '12 at 2:12
    
Pull the $x^2$ out of the denominator. –  Will Orrick Apr 13 '12 at 2:14
    
Sorry, I must be being dense (too many hours at the library today, no doubt), but then I have: $$\left(\dfrac{1 - \cos x}{x^2}\right) \left(\dfrac{2x^{3/2}}{3 \sqrt{\sin x} (x \cos x + \sin x)}\right)$$ If I'm not messing something up. I'm not sure I see that second term as an improvement. –  user23784 Apr 13 '12 at 2:23
    
Apportion the $x^{3/2}$ appropriately between the factors in the denominator and then evaluate the limits of those factors separately. –  Will Orrick Apr 13 '12 at 2:26
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yup. Just saw it as you posted that. Thanks, great solution. –  user23784 Apr 13 '12 at 2:26
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It might be appropriate to use the limits or infinitesimal properties of $\sin x$ at $x=0$

$${x- \sin x} \sim \frac{x^3}{3!}$$

Then

$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x - x}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^3}}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}}$$

and

$$\frac{1}{6}\mathop {\lim }\limits_{x \to {0^ + }} \frac{{{x^3}}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}{\left( {\mathop {\lim }\limits_{x \to {0^ + }} \frac{x}{{\sqrt {x\sin x} }}} \right)^3} = \frac{1}{6}{\left( {\mathop {\lim }\limits_{x \to {0^ + }} \sqrt {\frac{x}{{\sin x}}} } \right)^3}$$

Using the known limit $$\frac{x}{{\sin x}} \to 1$$

one has

$$\mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sin x - x}}{{{{\left( {x\sin x} \right)}^{\frac{3}{2}}}}} = \frac{1}{6}$$

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Yeah, this definitely works. I was hoping for a way to apply L'Hopital's directly. –  user23784 Apr 13 '12 at 2:09
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@rar L'Hopitals is overrated. –  Pedro Tamaroff Apr 13 '12 at 2:09
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First answer:

Try squaring the function first. The limit of $f^2$ will exist, and may be less messy via L'Hospital, although it will take more (twice as many) iterations. Then you'll need to decide which square root to take. But surely this $f$ is positive for small $x$.

Better answer: Assuming you know $\lim_{x\to0}\frac{\sin(x)}{x}=1$ (which you certainly do if LH is available), use that to your advantage to remove $\sin$ from the denominator:

$$ \begin{align} \lim_{x\to0^+}\frac{x-\sin x}{(x\,\sin x)^{3/2}}&=\lim_{x\to0^+}\frac{x-\sin x}{(x\,\sin x)^{3/2}}\cdot\left(\frac{\sin x}{x}\right)^{3/2}\\&=\lim_{x\to0^+}\frac{x-\sin x}{x^3} \end{align} $$

Now L'Hospital is a cake walk.

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Ah, yes that works very nicely as well. It all seems so obvious now... –  user23784 Apr 13 '12 at 2:28
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