Why should the numerator of this limit be quadratic?

Question

The limit is $$\frac{(a-2)x^3+(3+c)x^2+(b-3)x+2+d}{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}}=4$$

And the possible values of $a,b,c,d$ are to be found. This is an example question in my book and right off the bat it is said $a-2=0$ since the limit is finite, why is this so ? This is the only query I have with this limit.

Answer 1

The given limit can be written as $$\lim_{x\to\infty}\frac{a-2+\frac{3+c}{x}+\frac{b-3}{x^2}+\frac{2+d}{x^3}}{\sqrt{\frac{1}{x^2}+\frac{a}{x^3}+\frac3{x^4}+\frac b{x^5}+\frac2{x^6}}+\sqrt{\frac1{x^2}+\frac2{x^3}-\frac{c}{x^4}+\frac3{x^5}-\frac d{x^6}}}$$ The denominator converges to zero, so should the numerator for limit to exist finitely.

Answer 2

Because the result is a constant polynomial, the degree of the numerator must equal the degree of the denominator. The only way to accomplish this is to cancel out the $x^3$ term, which is true if $a-2 = 0$.

Answer 3

I'm going to offer an informal approach here; this can be made rigorous (and should!) but for getting a handle on questions like this it's often good to start informally.

The key piece of the informality is that polynomials are dominated by their leading term (as x goes to infinity). (There's a second-order informality, the linear binomial theorem, which says that $(1+t)^n\approx 1+nt$ as $t\to 0$, but that's not needed for the basic analysis here.)

How do these work here? Well, if we look at the denominators we have two terms each of the form $\sqrt{x^4+\mathrm{blah}+\mathrm{blah}+\mathrm{blah}}$. And since the 'blah's here are all smaller than $x^4$, they're irrelevant in the limit, so we can think of this as $\sqrt{x^4}$ plus some smaller stuff; in other words, $x^2$. So now your expression is $\dfrac{(a-2)x^3+\mathrm{smaller\ stuff}}{2x^2+\mathrm{smaller\ stuff}}$; this is what mandates $a=0$ for the limit to be finite. We can even go a step further by breaking down the numerator as $(a-2)x^3+(3+c)x^2+\mathrm{smaller\ stuff}$; then once we set $a=2$ to get a finite limit, we can see that the specific finite limit will be $\frac{3+c}2$, and adjust $c$ appropriately to make that equal the desired value.