<s>[INST] Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$ 
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!}+\frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-\frac{(x+x^3)^3}{3!}+\frac{(x+x^3)^5}{5!}-\frac{(x+x^3)^7}{7!}$
Now what do I do?
 [/INST] The limit in question is more easily solved by a little bit of algebra. Note that $$\frac{\sin(x+x^3)-x}{x^3}=\frac{\sin (x+x^3)-(x+x^3)}{(x+x^3)^3}\cdot (1+x^2)^3+1$$ Putting $t=x+x^3$ we can see that $t\to 0$ and hence the first factor $(\sin t - t) /t^3$ in first term tends to $-1/6$ and the desired limit is thus $(-1/6)+1=5/6$.
There is no need to find the Maclaurin series for $\sin(x+x^3)$, just the well known series for $\sin x $ is sufficient to tackle the problem. 
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