I am supposed to conjecture (estimate?) a value of the following limit using a Taylor series.
$$\lim_{x\to 0}\frac{\sin(x^2)-x^2}{x^6}$$
I am not sure though whether this is the right way? If someone could confirm, and if I am doing it wrong, point me in the correct direction to understand what to do.
$$\lim_{x\to 0}\frac{\left (x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\dots\right )-x^2}{x^6}=-\frac{1}{6}$$
Thanks
Put $t=x^2$. the limit will be $$\lim_{t\to 0}\frac {\sin (t)-t}{t^3}=$$
$$\lim_{t\to 0}\frac {t-\frac {t^3}{6}-t+t^3\epsilon(t)}{t^3}=\frac {-1}{6} $$