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?
 A: 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. 
A: Note that the formula for Maclaurin Series is
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$\sin(y)=\frac{0}{0!}y^0+\frac{1}{1!}y^1-\frac{0}{2!}y^2-\frac{1}{3!}y^3+\frac{0}{4!}y^4+...\\=y-\frac{y^3}{3!}+\frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $\sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=\sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)\cos(x+x^3)=1\\f^{(3)}(0)=6\cos(x+x^3)-18x(1+3x^2)(\sin(x+x^3))-(3x^2+1)(\cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$\sin(x+x^3)=x-\frac{5x^3}{6}.$$
Now we need to find $$\lim\limits_{x\to 0}\frac{\sin(x+x^3)-x}{x^3}.$$Plugging the series in for $\sin(x+x^3)$ we have
$$\lim\limits_{x\to 0}\frac{x-\frac{5x^3}{6}-x}{x^3}=\lim\limits_{x\to 0}\frac{\frac{5x^3}{6}}{x^3}=\lim\limits_{x\to 0}\frac{5x^3}{6x^3}=\frac56.$$
Hope this helps.
A: jnh's answer is correct, other than a sign error in the middle of it which is corrected by the end.
But you can do this without taking any derivatives by hand. The trick is to partially expand the binomial. You have:
$$\sin(x+x^3)=\sum_{n=0}^\infty \frac{(x+x^3)^{2n+1} (-1)^n}{(2n+1)!}$$
Considering the binomial theorem, the first term contributes a term of order $1$ and a term of order $3$, while the second term contributes a term of order $3$ and higher order terms. More generally the $n$th term contributes a term of order $2n+1$ and higher order terms, so going up to $n=1$ retrieves the first two terms of the desired series. 
The first two terms in the overall series will be order $1$ and $3$. The term of order $1$ can just be read off; it is just $x$. The term of order $3$ is a combination of the contribution from the $n=0$ term ($x^3$) and the contribution from the $n=1$ term, which by the binomial theorem is $\frac{-x^3}{6}$.
So we have $\sin(x+x^3)=x+\frac{5x^3}{6}+o \left ( x^3 \right )$. (The $o \left ( x^3 \right )$ is actually of order $5$, but this is not required to solve the problem.) Plugging this in we get
$$\frac{\sin(x+x^3)-x}{x^3}=\frac{\frac{5x^3}{6}+o \left ( x^3 \right )}{x^3} = \frac{5}{6} + o(1)$$
and so the desired limit is $\frac{5}{6}$. This approach of combining terms of the same order is common in perturbation theory problems in physics.
