$$\lim_{x\to0^+}\frac{1-\cos(x)}{x^2\sin(x)}$$
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.
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$\begingroup$Your limit is $+\infty$. Because $\dfrac{1-\cos x}{x^2} \to \dfrac{1}{2}$, and $\dfrac{1}{\sin x} \to +\infty$
$\endgroup$ $\begingroup$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 $x\to 0$.
$\endgroup$ $\begingroup$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: $$ \lim_{x\rightarrow 0^+}\dfrac{\cos\left(x\right)}{- x^2\,\sin\left(x\right)+2\,\sin\left(x\right)+4\,x\,\cos\left(x\right)} = +\infty $$
$\endgroup$ 0 $\begingroup$$$\lim\limits_{x\to +0}\frac{1-\cos x}{2\left(\frac{x}{2}\right)^2}\cdot\frac{1}{2\sin x}=\lim\limits_{x\to +0}\frac{2\sin^2\frac{x}{2}}{2\left(\frac{x}{2}\right)^2}\cdot\frac{1}{2\sin x},$$ now $\dfrac{\sin\frac{x}{2}}{\frac{x}{2}}\to 1$, ...
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