Tricky Limit question What is
$$\lim_{t \rightarrow 0} \frac{e^{t^3}-1-t^3}{\sin(t^2)-t^2}?$$
I used Hopital's rule, but it kept getting more complicated!

You would need to apply De l'Hôpital's Rule six (!) times before obtaining an answer.
From the Taylor expansions, we know $\sin(t^2) = t^2 - \dfrac{t^6}{3!} + O(t^{10})$ and $\exp(t^3) = 1 + t^3 + \dfrac{t^6}{2!} +  O(t^9)$.
Thus, the quotient becomes:
$$\lim_{t \to 0} \frac{e^{t^3}-1-t^3}{\sin(t^2)-t^2} = \lim_{t\to 0} \frac{t^6/2! + O(t^9)}{t^6/3! + O(t^{10})} = \frac{3!}{2!} = 3$$
