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