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?

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.
