The limit in question is:
$$\lim_{x\to 0} \frac{e^x \sin x - x(x+1)}{x^3}$$
The answer is supposedly $1/3$, and it is quite simple to find the limit using L'Hopital's rule. However, we are only permitted to use Taylor series formulas in class for solving this type of questions, and, by applying Taylor series to $e^x$ and $\sin x$, I don't seem to be able to get anything that can cancel with $x^3$, so I always end up getting
$$\frac{1}{x^n}$$
where $n$ is $1,2$ or $3$, and since $x$ approaches $0$, this renders the limit unsolvable (or incorrect) as I get multiple infinite values adding and subtracting one another.
I would be grateful if anyone could help me out with this.
Since the denominator is $x^3$, you have to at least expand the exponential and the sine function up to order three. Consider this:
$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3)$$
$$\sin(x) = x - \frac{x^3}{6} + o(x^3)$$
Thence
$$e^x \sin(x) = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + o(x^3)\right)\left(x - \frac{x^3}{6} + o(x^3)\right) = x + x^2 - \frac{x^3}{6} + \frac{x^3}{2} + o(x^3)$$
That is
$$e^x \sin(x) = x + x^2 + \frac{x^3}{3} + o(x^3)$$
Where we have trashed terms of order higher than $x^3$.
Then we have
$$\lim_{x\to 0} \frac{x + x^2 + \frac{x^3}{3} + o(x^3) - x^2 - x}{x^3} = \lim_{x\to 0} \frac{\frac{x^3}{3} + o(x^3)}{x^3} = \lim_{x\to 0} \frac{1}{3} + \underbrace{\lim_{x\to 0} \frac{o(x^3)}{x^3}}_{\to 0} = \frac{1}{3}$$