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$$\lim_{x\to0^+}\frac{1-\cos(x)}{x^2\sin(x)}$$

I keep running in circles using the L'Hospital rule. After the third time applying it I got 0 but this isnt true from the graph. I can see it goes to +ve infinity.

Please let me know if anyone has an elegant solution to this lengthy problem.

Joel
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Sherlock Homies
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  • Remember that you can only apply L'Hôpital when it's in the form $\frac00$ (or $\frac\infty\infty$). If either the numerator or the denominator are not $0$ (or alternatively $\infty$), it will not work. – Akiva Weinberger Jun 01 '15 at 20:12
  • The denominator can alone go to $\pm \infty$, don't necessarily require $\frac{\infty}{\infty}$. – Sudarsan Jun 01 '15 at 20:14
  • @OP See Lykos's answer for the fact that L'Hospital's works too. We clearly have a $\frac{0}{0}$ case and the derivative of the denominator is non-zero for a small interval near $0$ [the two conditions needed to apply L'Hospital's]. – Sudarsan Jun 01 '15 at 20:27
  • @ Sudarsan hey how about Alexeys answer? can you guys confirm that? – Sherlock Homies Jun 01 '15 at 21:04

4 Answers4

4

Your limit is $+\infty$. Because $\dfrac{1-\cos x}{x^2} \to \dfrac{1}{2}$, and $\dfrac{1}{\sin x} \to +\infty$

DeepSea
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The limit is indeed $\infty$. Recall that $1-\cos x=\frac12 x^2+O(x^4)$ and $x^2\sin x=x^3+O(x^5)$.

Thus,

$$\frac{1-\cos x}{x^2 \sin x}=\frac{1}{2x}+O(x)\to \infty$$

as $x\to 0$.

Mark Viola
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2

the first application of the rule gives: $$ \lim_{x\rightarrow 0^+}\dfrac{\sin\left(x\right)}{ 2\,x\,\sin\left(x\right) + x^2 \,\cos\left(x\right)} $$ the second application of the rule gives: $$ \lim_{x\rightarrow 0^+}\dfrac{\cos\left(x\right)}{- x^2\,\sin\left(x\right)+2\,\sin\left(x\right)+4\,x\,\cos\left(x\right)} = +\infty $$

Wolfgang Brehm
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$$\lim\limits_{x\to +0}\frac{1-\cos x}{2\left(\frac{x}{2}\right)^2}\cdot\frac{1}{2\sin x}=\lim\limits_{x\to +0}\frac{2\sin^2\frac{x}{2}}{2\left(\frac{x}{2}\right)^2}\cdot\frac{1}{2\sin x},$$ now $\dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\to 1$, ...

Alexey Burdin
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