How to find the finite limit of this function? 
Let $f(x) = \dfrac{1-\cos \{x\}}{(x^4 + ax^3 +bx^2 +cx)^2}$. If $l= \lim_{x\to 1^+}f(x), m = \lim_{x\to 2^+}f(x) $ and $n= \lim_{x\to 3^+}f(x),$ where $l,m$ and $n$ non-zero finite then:
$a+b+c=? $
$l+m+n=?$ 
$\lim_{x\to 0^+}f(x)=? $
  where {} denotes the fractional part function. 

The trouble I am facing with this question is that for $n^+$ for $n \in \mathbb N$, the numerator is turning out to be $0$ as $\{x\}= 0$ and $\cos 0 =1$ and the deominator is finite. 
I even tried the taylor expansions of $\cos \{x\}$ but that didn't help.  I don't need the full solution, just want a hint to be able to proceed. 
 A: Try breaking the fraction part function into different intervals, like $(1,2)$ etc. For $x\in (1,2) $ we have $\{x\} = x-1$ and so
$$\begin{align}
l=\lim_{x\to1^+} \frac{1-\cos\{x\}}{(x^4+ax^3+bx^2+cx)^2} &= \lim_{x\to1^+}\frac{2\sin^{2}(\frac{x-1}{2})}{(\frac{x-1}{2})^2}\cdot \frac{(\frac{x-1}{2})^2}{(x^4+ax^3+bx^2+cx)^2}\\
&= \lim_{x\to1^+}\frac{(x-1)^2}{2(x^4+ax^3+bx^2+cx)^2}
\end{align}$$
For the limit to be nonzero finite, we need $0/0$ form, so we need $(x=1)$ to be root of denominator quartic. Similar analysis for limits $m,n$ gives us three roots of quartic on denominator, which are $x=1,2,3$. 
The fourth root is obviously $x=0$. Call the denominator quartic as $P(x)=x^4+ax^3+bx^2+cx$. We have shown $P(x) = x(x-1)(x-2)(x-3)$. So now finding limits $l,m,n$ is simple:


*

*$l= \frac{1}{2 \cdot 1\cdot 4\cdot 1} =  1/8$

*$m= \frac{1}{2 \cdot 1\cdot 1\cdot 4} = 1/8$

*$n= \frac{1}{2 \cdot 1\cdot 4\cdot 9} = 1/72$


From here we get $l+m+n = \frac{19}{72}$
For finding $a,b,c$, we have $P(x)=x^4+ax^3+bx^2+cx = x(x-1)(x-2)(x-3)$ so $P(1) = a+b+c+1=0$ or $a+b+c=-1$.
