Limit of an unknown cubic expression Let $f(x)=ax^3+bx^2+cx+d$ and $g(x)=x^2+x-2$. If $$\lim_{x \to 1}\frac{f(x)}{g(x)}=1$$ and $$\lim_{x \to -2}\frac{f(x)}{g(x)}=4$$ then find the value of $$\frac{c^2+d^2}{a^2+b^2}$$ Since the denominator is tending to $0$ in both cases, the numerator should also tend to $0$, in order to get indeterminate form. But it led to more and more equations.
 A: What about? $$f(x)=-x^3+x^2+4x-4$$
So:
$$\frac{c^2+d^2}{a^2+b^2}=32/2=16$$

Let us make ansatz that $$f(x)=a(x+\alpha)(x-1)(x+2)$$
So:
$$a(1+\alpha)=1,a(\alpha-2)=4\implies a=-1,\alpha=-2$$
so:
$$f(x)=-(x-1)(x+2)(x-2)=-x^3+x^2+4x-4$$
A: You pointed out that the numerator should also tend to $0$ when $x \to 1$ or $x \to -2$. So, let's make our polynomial
$$f(x) = (ax + b)(x - 1)(x + 2).$$
Now $$\lim_{x \to 1}\frac{f(x)}{g(x)} = \frac{(a+b)}{3} = 1,$$ and $a + b = 3$. Also,
$$\lim_{x \to -1}\frac{f(x)}{g(x)} = \frac{(b - 2a)}{-3} = 4,$$ and $b - 2a = -12$. Now you've got two unknown parameters and two linear equations, so that shouldn't be an issue to find $a$ and $b$. Once you have them, you can multiply out the polynomial to find $c$ and $d$, as well as the final fraction.
Maybe a bit inelegant, but it'll get the job done!
A: Use polynomials long division to have the form
$$ f(x) = ax^3+bx^2+cx+d=(( 3a-b+c )x-2a+2b+d)(x^2+x-2)$$

$$ \implies f(x) = (( 3a-b+c )x-2a+2b+d) g(x) $$

which simplifies evaluating the limit and the problem becomes straightforward.
