Line that matches (x³ − 1)/(x² + 1) at infinity
Find constants $a$ and $b$ such that: $$ \lim_{x \to \infty} \left(ax+b - \frac{x^3-1}{x^2+1}\right) = 0. $$
Explanation of Method
Since the numerator $x^3-1$ has degree one higher than the denominator $x^2+1$, the rational function $\frac{x^3-1}{x^2+1}$ behaves like a linear function $ax+b$ for large $x$, plus a remainder that vanishes at infinity — exactly the definition of an oblique (slant) asymptote. Polynomial long division of $x^3-1$ by $x^2+1$ makes this precise: it produces the linear quotient directly, together with an explicit remainder term whose vanishing as $x\to\infty$ can be checked rigorously (rather than merely approximated).
- Divide $x^3-1$ by $x^2+1$ using polynomial long division.
- Identify the quotient as the candidate asymptote $ax+b$.
- Verify that the remainder term vanishes as $x\to\infty$, confirming the condition holds for these values of $a,b$ (and no others).
Step 1 · Perform polynomial long division
Divide $x^3-1$ by $x^2+1$: since $x^3 = x\cdot(x^2+1) - x$, we have $x^3-1 = x(x^2+1) - x - 1$. So:
$$ \frac{x^3-1}{x^2+1} = x + \frac{-x-1}{x^2+1} = x - \frac{x+1}{x^2+1}. $$
Step 2 · Identify the candidate asymptote
This expresses the original function exactly as the linear part $x$ plus a remainder term $-\dfrac{x+1}{x^2+1}$ that has a strictly smaller degree in the numerator ($1$) than the denominator ($2$). This suggests the asymptote is $ax+b$ with $a=1$, $b=0$.
Step 3 · Verify the remainder vanishes
Substituting $a=1,\,b=0$ into the original limit expression:
$$ ax+b-\frac{x^3-1}{x^2+1} = x - \left(x-\frac{x+1}{x^2+1}\right) = \frac{x+1}{x^2+1}. $$
Since the numerator has degree $1$ and the denominator has degree $2$, this ratio tends to $0$ as $x\to\infty$ (dividing numerator and denominator by $x^2$ gives $\frac{1/x+1/x^2}{1+1/x^2}\to\frac{0+0}{1+0}=0$). So indeed:
$$ \lim_{x\to\infty}\left(x-\frac{x^3-1}{x^2+1}\right) = \lim_{x\to\infty}\frac{x+1}{x^2+1} = 0. $$
Moreover, these are the only values of $a,b$ that work: if $a\neq1$, the difference $ax+b-\frac{x^3-1}{x^2+1} = (a-1)x+b+\frac{x+1}{x^2+1}$ would contain a nonvanishing linear term $(a-1)x\to\pm\infty$; and if $a=1$ but $b\neq0$, the difference would tend to the nonzero constant $b$ instead of $0$.
Summary Table
| Quantity | Exact expression | Limit as $x\to\infty$ |
|---|---|---|
| $\dfrac{x^3-1}{x^2+1}$ | $x-\dfrac{x+1}{x^2+1}$ | — |
| $x-\dfrac{x^3-1}{x^2+1}$ | $\dfrac{x+1}{x^2+1}$ | $0$ |
Additional Notes
- Using exact polynomial long division (rather than the informal shortcut of just dropping the "$-1$" and "$+1$" and approximating $\frac{x^3-1}{x^2+1}\approx x$) makes it possible to verify precisely, not just plausibly, that the remainder term vanishes — and to see exactly how fast it does (like $\frac1x$ for large $x$).
- This is the general method for finding an oblique asymptote of any rational function where the numerator's degree exceeds the denominator's by exactly $1$: perform the division, and the quotient (ignoring the remainder) is the asymptote.
- Since the argument in Step 3 shows both conditions ($a=1$ and $b=0$) are forced — not just sufficient — this pins down the unique straight line that the graph of $\frac{x^3-1}{x^2+1}$ approaches as $x\to\infty$.
Result
The required constants are: