Finding $(a, b, c)$ for which $\lim_{x\to0^+}\frac{\arctan\left(x+\frac{a}{x}\right)+bx+c}{x^5}$ exists, and the corresponding limiting values 
Find all possible values of the triple $(a, b, c)$ of constants for which the limit 
  $$\displaystyle \lim_{x\to0^+} \frac{\arctan\left(x+\frac{a}{x}\right)+bx+c}{x^5}$$ 
  exists and find the limit for each of these values of $(a, b, c)$.

So, I’ve been trying to solve this problem by using L’Hospital’s Rule over and over but it doesn’t lead me anywhere? What should I do? Can somebody lead me?
 A: Hint:
Applying L'Hospital, we have $$\frac{\dfrac{1-\dfrac a{x^2}}{1+\left(x+\dfrac ax\right)^2}+b}{5x^4}=\frac{\dfrac{x^2-a}{x^2+\left(x^2+a\right)^2}+b}{5x^4}=\frac{bx^4+(2ab+b+1)x^2+ba^2-a}{5x^4\left(x^2+\left(x^2+a\right)^2\right)}.$$
The existence of the limit requires that the quadratic and constant terms vanish, $a=0,b=-1$ or $ab=1,b=-3$. But these conditions might not suffice...
A: For the limit to possibly exist, it needs to be of the form $\frac 00$. Otherwise, it's of the form $\frac{\text{non-zero number}}{0^+}$ which diverges to $\infty$ or $-\infty$. This is the case when $a\ne 0$ and $c\ne \pm\frac \pi2$. Note that $\arctan\left(x+\frac ax\right)\to \pm\frac \pi2$ depending on whether $a>0$ or $a<0$. So you have 2 cases:
Case 1: $a=0$. Then $c=0$ (to make it of the form $0/0$). Applying L'Hopital's rule, we have:
$$\lim_{x\to0^+}\frac{\frac{1}{1+x^2}+b}{5x^4}=\lim_{x\to 0^+}\frac{bx^2+b+1}{5x^4(x^2+1)} $$
You need $b=-1$. But after making cancellations, the limit doesn't exist.
Case 2: $c=\frac \pi 2$ with $a<0$ or $c=-\frac \pi 2$ with $a>0$. Applying L'Hopital's rule and simplifying:
$$\frac{1}{a^2}\lim_{x\to0^+}\frac{bx^4+(2ab+b+1)x^2+a^2b-a}{5x^4} $$
And you need $a^2b-a=0\Rightarrow b=\frac{1}{a}$ and $2ab+b+1=\frac{1}{a}+3=0\Rightarrow a=-\frac{1}{3}$, so $b=-3$ and $c=\frac \pi2$ since $a<0$. After making cancellations, we see the limit exists for those values.
