<s>[INST] Limit as $x$ tends to zero of $\frac{\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}$ 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. 
 [/INST] 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 $$ \lim_{x \to 0}\left(\frac {\csc(x)}{x^3} - \frac{\sinh(x)}{x^5}\right) = \frac1{90}. $$
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