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How to find $$\lim_{x\to 0}\frac{\sinh x - \sin x}{x\sin^2 x}?$$ Tried by L'Hospital rule but the process does not stop.

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The process does stop after three iterations. But to ease the computation, we first transform with

$$\frac{\sinh x-\sin x}{x\sin^2x}= \frac{\sinh x-\sin x}{x^3}\frac{x^2}{\sin^2x}$$ where the second factor is known to converge to $1^2$.

Then $$\frac{\sinh x-\sin x}{x^3}\to\frac{\cosh x-\cos x}{2x^2}\to\frac{\sinh x+\sin x}{6x}\to\frac{\cosh x+\cos }6$$

gives the solution

$$\frac 26.$$


Anyway,

$$x\sin^2x\to\sin^2x+2x\sin x\cos x\to4\sin x\cos x+2x(\cos^2x-\sin^2x)\to\\ 6(\cos^2x-\sin^2)-8x\sin x\cos x$$ also leads to the denominator $6$.

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Use the fact that$$\sinh x-\sin x=\frac{x^3}{3}+o(x^3)$$and that$$x\sin^2x=x^3+o(x^3).$$

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