The value of a, b and c for this integral 
Find out the value of $a, b$ and $c$, for which
$$ \lim _{x\to 0}\:\:\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\frac{1}{5} $$

My first thought was to separate the integral into multiple ones because i thought that it is pretty clear that each integral is a common one and a,b,c are constants and can be taken out . After I solved the integral, that's where I had the problem. I couldn't do the limit of the result that I've got in order to find out the value of $a,b$ and $c$ .
 A: Note that, as $x\to 0$, 
$$I=\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\to0$$and $x^5\to 0$. Thus, we can apply L-hop rule to get 
$$ L=\lim _{x\to 0}\:\:\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\lim_{x\to 0}\frac{a+b\cos(x)+c\cos(2x)}{5x^4} $$
Clearly, for this limit to converge, the numerator of the fraction should be zero at $x=0$ (why?). Thus $a+b\cos(0)+c\cos(0)=0\implies\boxed{a+b+c=0}$.
Since, again the fraction at $x=0$ is of the form $0/0$, we can again apply L-hop rule to get
$$L=\lim_{x\to 0}\frac{-b\sin(x)-2c\sin(2x)}{20x^3}$$
Applying the  similar argument again, we get nothing special because at $x=0$, the numerator is zero for any $b,c\in\mathbb{C}$. Thus, the fraction is again $0/0$ form and hence we can again apply L-hop rule to get 
$$L=\lim_{x\to 0}\frac{-b\cos(x)-4c\cos(2x)}{60x^2}$$
This time, again using the same argument we get, $-b-4c=0\implies \boxed{b=-4c}$. Now, I guess you know how to continue further (just apply L-hop two more times to get the third equation in $b,c$ and this is just enough (why?)) .
Note: When you apply the last L-hop, don't forget to use the information that $L=0.2$
A: Hint:
As the integral is $\; ax+b\sin x+\frac c2\sin 2x$, plug in the expansion of $\sin x$ at order $5$:
$$\sin x=x-\tfrac13x^3+\tfrac1{120} x^5+o(x^5)$$
to obtain a linear system in $a, b,c$.
A: It certainly looks like direct integration should work.  
$\int_0^x (a+ b \cos(t)+ \cos(2t))dt= ax+ b \sin(x)+ \frac{1}{2} \sin(2x)$
so $\frac{1}{x}\int_0^x (a+ b \cos(t)+ \cos(2t))dt= a+ b\frac{\sin(x)}{x}+ \frac{\sin(2x)}{2x}$
The limit of $\frac{\sin(\theta)}{\theta}$ as $\theta$ goes to 0 is a standard one.
A: $$
\lim _{x\to 0}\frac{1}{x^5}\left(\int _0^x\:\left(a+b \cos\left(t\right)+c \cos\left(2t\right)\right)dt\right)=\lim _{x\to 0}\frac{a x+\sin (x) (b+c \cos (x))}{x^5}
$$
If this is to be equal to $1/5$, then you must have
$$
a x+\sin (x) (b+c \cos (x))=\frac15x^5+\mathcal O(x^6)
$$
and so it's just a matter of Taylor expanding:
$$
a x+\sin (x) (b+c \cos (x))=(a+b+c)x- \frac16(b+4 c)x^3+\frac{1}{120} (b+16 c)x^5 +\mathcal O(x^7)
$$
Can you take it from here?
A: Let us call the limit $L$.
Since both $x^5$ and the integral tend to zero as $x\to0$, l'Hopital's rule and the fundamental theorem of calculus can be used to find
$$
L
=
\lim _{x\to 0}\frac{1}{5x^4}\left(a+b \cos\left(x\right)+c \cos\left(2x\right)\right).
$$
There are several ways one can go from here.
You can keep applying l'Hopital, but be careful to make sure that the limits of both the numerator and the denominator are zero.
I will use Taylor series instead.
Since we're dividing by $x^4$, we need to expand the numerator to that order.
We have
$$
\cos(x)
=
1-\frac12x^2+\frac1{24}x^4+O(x^5),
$$
so
$$
a+b\cos(x)+c\cos(2x)
=
[a+b+c]
-
\left[\frac12b+2c\right]x^2
+
\left[\frac1{24}b+\frac23c\right]x^4
+
O(x^5).
$$
Therefore
$$
L
=
\lim{x\to0}
\frac15
\left(
[a+b+c]x^{-4}
-
\left[\frac12b+2c\right]x^{-2}
+
\left[\frac1{24}b+\frac23c\right]
+
O(x)
\right).
$$
So, to have the limit exist as a real number in the first place, you get the conditions $a+b+c=0$ and $\frac12b+2c=0$.
To have the correct value, you get a third equation.
This linear system can be solved for $a,b,c$ with standard tools.
