Find $a^3-b^2$ where a,b satisfies:
$\lim \limits_{x\to\infty} ((x^2+1)/(x+1) - ax-b) = 0.$
I tried equating it to $0$ I.e. \begin{align}&\left(\frac{x^2+1}{x+1} - ax-b\right)= 0 \\ & \implies \frac{x^2+1}{x+1} = ax+b \\ & \implies x^2+1= ax^2 + ax + bx + b \end{align} Then I compared the coefficients Which gave \begin{align}a&= 1,\\ a+b &= 0,\\ b &= 1, \end{align} But if $a=b=1$ then how can $a+b =0$ am confused at this point.
Can't figure whats wrong with my steps
Since $\frac{x^2+1}{x+1}$ is not a polynomial, we can't equate it directly to a linear polynomial.
However, as $x$ goes to $\infty$, it behaves as if it is linear. As you can see from the picture, though it is not a linear function, as $x$ grows bigger, it behaves almost linearly.
Since
$$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1},$$
we have
\begin{align}\lim_{x \to \infty} \left[\frac{x^2+1}{x+1}-ax-b\right]&=\lim_{x \to \infty} \left[(1-a)x+(-1-b)+\frac{2}{x-1}\right] \\ &= \lim_{x \to \infty}(1-a)x+(-1-b) \end{align}
Note that $\lim_{x \to \infty} \frac{2}{x-1}=0$.
You should be able to read off the value of $a$ and $b$ from the discussion above.