I'm trying to use Taylor series to compute the following limit. I'm not the best with the following method so I'm trying my best to practice with it, and would be happy to receive feedback and support.
$\lim_{x \to 0}\frac{1-\cos^2(x)}{x^2\cdot \sin(x)}$
Given that :
$1-\cos^2(x) = \sin^2(x)$
The Taylor series expansion of $\sin^2(x)$ is:
$x^2 - \frac{x^4}{3} + \frac{(2 x^6)}{45} - \frac{x^8}{315} + O(x^9)$
And the Taylor series expansion for $x^2\sin(x)$ is:
$x^3 - \frac{x^5}{6} + \frac{x^7}{120} + O(x^9)$
Putting all of this together:
$$\frac{x^2 - \frac{x^4}{3} + \frac{(2 x^6)}{45} - \frac{x^8}{315} + O(x^9)}{x^3 - \frac{x^5}{6} + \frac{x^7}{120} + O(x^9)} \implies \frac{8}{21}?$$
I used wolfram to calculate the fraction as it was too large for me to compute. Though, have I approached this correctly, and was I right in picking the following fraction $\frac{8}{21}?$
You have, near $0$,\begin{align}\lim_{x\to0}\frac{1-\cos^2(x)}{x^2\sin(x)}&=\lim_{x\to0}\frac{x^2-\frac13x^4+\cdots}{x^3-\frac16x^5+\cdots}\\&=\lim_{x\to0}\frac{1-\frac13x^2+\cdots}{x-\frac16x^3+\cdots}\end{align}and this limit does not exist, since$$\lim_{x\to0^+}\frac{1-\frac13x^2+\cdots}{x-\frac16x^3+\cdots}=\infty\quad\text{and}\quad\lim_{x\to0^-}\frac{1-\frac13x^2+\cdots}{x-\frac16x^3+\cdots}=-\infty.$$