<s>[INST] First four nonzero terms of the McLaurin expansion of $\frac{xe^x}{\sin x}$ at $x_0=0$ 
Define if necessary the given function so as to be continuous at $x_0=0$ and find the first four nonzero terms of its MacLaurin series.
  $$ \frac{xe^x}{\sin x}$$

Given $f(x)= \frac{h(x)}{g(x)}$ where $h(x)= xe^x$, and $g(x)=\sin x$
We have the following MacLaurin series
$$f(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x +a_2 x^2 + a_3 x^3 +a_4 x^4 + ...$$
$$g(x) = \sum_{n=0}^{\infty} b_n x^n = 0 + x+ 0 -\frac{1}{6} x^3 + 0 + \frac{1}{120} x^5 +...$$
$$h(x) = xe^x = \sum_{n=0}^{\infty} c_n x^n =   \sum_{n=0}^{\infty} \frac{n}{(n)!}x^n = 0 + x + x^2 + \frac{1}{2} x^3 + \frac{1}{6} x^4  + \frac{1}{24} x^5$$
.
The Maclaurin coefficients of $f(x)g(x)$
$$ c_0 = a_0b_0 = 0$$
$$ c_1 = a_0b_1 + a_1b_0 = a_0$$
$$ c_2 = a_0b_2 + a_1b_1 +a_2b_0 = a_1 $$
$$ c_3 = a_0b_3 +\cdots a_3b_0 = - \frac{1}{6} a_0 + a_2$$
$$ c_4 = a_0b_4 +\cdots + a_4b_0 = - \frac{1}{6} a_1 + a_3 $$
$$ \implies f(x)g(x) = a_0 x^1 + a_1 x^2 + \left( - \frac{1}{6} a_0 + a_2\right)x^3 + \left(- \frac{1}{6} a_1 + a_3\right) x^4 + \cdots$$
.
Equating coefficients of $f(x)g(x)$ with coefficients of $h(x)$
$$ c_0 = 0$$
$$ c_1 =  a_0  = 1 $$
$$ c_2  = a_1 = 1 $$
$$ c_3  = - \frac{1}{6} a_0 + a_2 = 1/2 \implies a_2 = 2/3 $$
$$ c_4  = - \frac{1}{6} a_1 + a_3 = 1/6 \implies a_3 = 1/3 $$
.
it follows that the first 4 nonzero terms are:
$$f(x) = \frac{x e^x}{\sin(x)} = 1 + x + \frac{2}{3} x^2 + \frac{1}{3} x^3$$
.
I would like to know if I am  going in the right direction? 
My main question regards the original question "Define if necessary the given function so as to be continuous at $x_0=0$" . I am not sure how to handle this question, if a definition is needed how? and why? if a definition is not needed how? and why?
Thx for your input/help.  
 [/INST] I didn't check the computations but, yes, you are doing it right.
As for the other question, note that the expression $\frac{xe^x}{\sin x}$ is undefined if $x=0$. In order to make $f$ continuous at $0$; you must define $f(0)$ as $\lim_{x\to0}\frac{xe^x}{\sin x}$, which is equal to $1$.
</s>