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This exercise $$\lim_{x\to 0+}\left[\frac{1}{x^\frac{3}{2}}-\frac{1}{x^\frac{1}{2}\sin(x)}\right]$$ was in my calculus III test at college, after trying hard to solve it I was not really able to do so, I know intuitively that this limit equals 0 but did not find an appropriate way to prove it, as the limit gives an indeterminate form $\infty - \infty$. I worked out the fractions and got $$\lim_{x\to 0+}\frac{x^\frac{1}{2}\sin(x) - x^\frac{3}{2} }{x^2\sin(x)}$$ which gives me an indeterminate form $\frac{0}{0}$, then using L'hopitals rule, the limit starts getting uglier as more indeterminate forms keep showing up. My college test is over, but, still I want to find out how to solve this limit, any help would be highly appreciated.

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  • $\begingroup$ Are you sure you correctly recall it? The reason I ask is that it seems to me that the second term is essentially $1/x^{3/2}$, which blows up faster than the first term, so they can't cancel... $\endgroup$ – paul garrett Jul 20 at 20:03
  • $\begingroup$ I'm sorry I had written one of the fractions incorrectly, question edited, now that is the real exercise $\endgroup$ – Sneyder Angulo Jul 20 at 20:05
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I wouldn't rationalise the denominator. Instead, stick with $$\frac{\sin x - x}{x^{\frac{3}{2}}\sin x}.$$ Applying L'Hopital's rule, \begin{align*} \frac{\cos x - 1}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x} &= \frac{(\cos x - 1)(\cos x + 1)}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\ &= \frac{-\sin^2 x}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\ &= \frac{-x^{\frac{1}{2}}\frac{\sin^2 x}{x^2}}{\left(\frac{3}{2}\frac{\sin x}{x} + \cos x\right)(\cos x + 1)} \\ &\to \frac{-0 \cdot 1}{\left(\frac{3}{2} \cdot 1 + 1\right)(1 + 1)} = 0. \end{align*}

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  • $\begingroup$ oh boy, this is what I was looking for, thank you very much. $\endgroup$ – Sneyder Angulo Jul 20 at 21:16
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Hint. By the Taylor series expansion, as $x \to 0$, one gets $$ \sin x=x-\frac{x^3}6+O(x^5) $$ giving $$ \frac1{\sin x}=\frac1x+\frac{x}6+O(x^3) $$ and, as $x \to 0^+$, $$ \frac1{x^{1/2}\sin x}=\frac1{x^{3/2}}+\frac{\sqrt{x}}6+O(x^{5/2}) $$ Hope you can take it from here.

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  • $\begingroup$ Hi, we had not been given Taylor series at college yet, so I guess there has to be another(simpler) way to solve it, anyways, thanks for your time and effort ;) $\endgroup$ – Sneyder Angulo Jul 20 at 20:10

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