Taylor Expansion of the first four non-zero terms $$f(x) = \sqrt{1+x}\sin(4x)$$
Okay, so at this point, I am able to find the answer to these kinds of problems when the outside term is x to the power of a whole number, but this is two terms added together then to the 1/2 power.
The sin expansion looks like this:

$$\sin(x) = x  - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$

and the sin expansion of sin(4x), looks like this: (I'm pretty sure)

$$\sin(4x) = 4x  - \frac{4x^3}{3!} + \frac{4x^5}{5!} - \frac{4x^7}{7!} + \cdots$$

but, what does this look like:

$$\sqrt{1+x}\sin(4x) = \cdots$$


Near $0$, you have$$\sin(4x)=4x-\frac{4^3}{3!}x^3+\cdots$$and$$\sqrt{1+x}=(1+x)^{1/2}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots,$$and therefore$$\sqrt{1+x}\sin(4x)=4x+2x^2-\frac{67x^3}6-\frac{61x^4}{12}+\cdots$$The RHS of this equality is what one gets after ignoring the terms of the product$$\left(4x-\frac{4^3}{3!}x^3\right)\times\left(1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}\right)$$whose degree is greater than $4$.
