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How would I find the $$\lim_{x\to0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)$$ The only way I know how to do this is with l'hopitals rule but I don't see it helping here as we have x's in our denominator.

Gregory Grant
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Goods
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    You can write the difference as a single quotient $f(x)/g(x)$ with denominator $x^5$. If the hypotheses of l’Hôpital’s rule apply to the single quotient as $x\to0$ (they will, but you should check), you’re good to go. It’s very tedious, since you have to apply the rule multiple times and check that the numerator (yuck) and denominator (simple) approach zero as $x$ approaches zero each time, but if you’re meticulous, you’ll get the right answer. – Steve Kass May 29 '15 at 19:24
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    Hint: $~\csc x=\dfrac1{\sin x}\simeq\dfrac1{x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}}\quad$ and $\quad\sinh x\simeq x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}~.~$ The reason for expanding until $x^5$ is due to the fact that terms up till $x^5$ appear in the expression of the limit. – Lucian May 29 '15 at 20:52

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you have $$ \left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right)= \frac{2x^2 - (e^x - e^{-x})\sin x}{2x^5\sin x}$$ we will look at maclaurin expansion of $$\begin{align} (e^x - e^{-x})\sin x - 2x^2 &=2\left(x+ \frac16 x^3+\frac1{120}x^5 + \cdots\right)\left(x - \frac16x^3 + \frac1{120}x^5+\cdots\right)-2x^2 \\ &=2\left(x^2+\left(\frac1{60} - \frac1{36}\right)x^6+ \cdots \right)-2x^2 \\ &= -\frac1{45}x^6 + \cdots\end{align} $$

therefore $$ \lim_{x \to 0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right) = \frac1{90}. $$

abel
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I'm not sure what you mean about L'Hospital's not helping because of $x$ in the denominator. As long as your limit satisfies one of the indeterminate forms, L'Hospital's rule is fair game. You may have to coax your equation to "appear" in indeterminate form. For example, write $$\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5} = \frac {x^2- \sin(x)\sinh(x)}{x^5\sin(x)}$$ It is now clear that $$\lim_{x \to 0} \left(\frac {x^2- \sin(x)\sinh(x)}{x^5\sin(x)}\right)$$ satisfies the $\frac{0}{0}$ indeterminate form.

graydad
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