Finding constant of limit question If k and b are constant such that,
$$\lim_{x \to \infty} (kx + b - \frac{x^3+1}{x^2+1}) = 0$$
Find the values of k and b.
From what I learn that if the degree of numerator is higher than denominator, then limit to infinity should also be infinity, not 0. And just from that, I'm stuck on this question.
 A: $$kx + b - \frac{x^3+1}{x^2+1}=\frac{kx^3+kx+bx^2+b-x^3-1}{x^2+1}=\\
=\frac{x^3(k-1)}{x^2+1}+\frac{b(x^2+1)}{x^2+1}+\frac{kx-1}{x^2+1}$$
taking $k=1$, I am sure, you easily say $b$.
A: $kx + b - \frac{x^3+1}{x^2+1}=(k-1)x+b+\frac {x-1}{x^{2}+1}$. The lst term tends to $0$ so the  whole thing tends to $0$ if and only if $x(k-1)+b \to 0$ Obviously, this means $k=1$ and $b=0$.
A: I will write the informal idea. If you want to find the $\displaystyle \lim_{x\to +\infty}\frac{x^3+1}{x^2+1}$,  then indeed the limit is $+\infty$ as you point out. However,  you have the expression $\displaystyle kx+b-\frac{x^3+1}{x^2+1}$,  then the term $kx$ can change all that previous history. Because we can write it as $$kx+b-\frac{x^3+1}{x^2+1}=b+x(k-\frac{x^3+1}{x(x^2+1)})$$
The last expression can be written as
$$b+x(k-\frac{1+\frac{1}{x^3}}{1+\frac{1}{x^2}})$$
But, notice that $$\lim_{x\to +\infty}\frac{1+\frac{1}{x^3}}{1+\frac{1}{x^2}}=1$$
This suggests how to take the values of $k$ and $b$. Of course,  then $(k,b)=(1,0)$ works.
