Limit of (x³ − 2x − 1)/(x^5 − 2x − 1)
Evaluate the limit: $$ L = \lim_{x \to -1} \frac{x^3 - 2x - 1}{x^5 - 2x - 1}. $$
Explanation of Method
Direct substitution of $x=-1$ makes both the numerator and denominator vanish, giving $\dfrac{0}{0}$. Since both are polynomials and $x=-1$ is a root of each, the Factor Theorem guarantees that $(x+1)$ divides both exactly. Dividing it out of each polynomial (by polynomial long division or synthetic division) produces a simplified fraction that is continuous at $x=-1$, so the limit can be found by direct substitution afterward.
- Confirm that $x=-1$ is a root of both the numerator and the denominator.
- Divide each polynomial by $(x+1)$ using synthetic division.
- Substitute $x=-1$ into the simplified quotient of polynomials.
Step 1 · Confirm the indeterminate form
Substituting $x=-1$ into the numerator:
$$ (-1)^3 - 2(-1) - 1 = -1+2-1 = 0. $$
Substituting $x=-1$ into the denominator:
$$ (-1)^5 - 2(-1) - 1 = -1+2-1 = 0. $$
Since both vanish, $x=-1$ is a common root, and by the Factor Theorem $(x+1)$ divides both $x^3-2x-1$ and $x^5-2x-1$ exactly.
Step 2 · Divide out the common factor $(x+1)$
Dividing the numerator $x^3-2x-1$ by $(x+1)$ using synthetic division with root $-1$ (coefficients $1,\,0,\,-2,\,-1$) gives quotient $x^2-x-1$ with remainder $0$:
$$ x^3 - 2x - 1 = (x+1)(x^2-x-1). $$
Dividing the denominator $x^5-2x-1$ by $(x+1)$ using synthetic division with root $-1$ (coefficients $1,\,0,\,0,\,0,\,-2,\,-1$) gives quotient $x^4-x^3+x^2-x-1$ with remainder $0$:
$$ x^5 - 2x - 1 = (x+1)\left(x^4-x^3+x^2-x-1\right). $$
So, for $x \neq -1$:
$$ \frac{x^3-2x-1}{x^5-2x-1} = \frac{(x+1)(x^2-x-1)}{(x+1)(x^4-x^3+x^2-x-1)} = \frac{x^2-x-1}{x^4-x^3+x^2-x-1}. $$
Step 3 · Substitute $x=-1$ into the simplified quotient
Evaluating the numerator of the simplified fraction at $x=-1$:
$$ (-1)^2-(-1)-1 = 1+1-1 = 1. $$
Evaluating the denominator at $x=-1$:
$$ (-1)^4-(-1)^3+(-1)^2-(-1)-1 = 1+1+1+1-1 = 3. $$
So:
$$ L = \frac{1}{3}. $$
Summary Table
| Polynomial | Factored form | Value at $x=-1$ (reduced) |
|---|---|---|
| $x^3-2x-1$ | $(x+1)(x^2-x-1)$ | $1$ |
| $x^5-2x-1$ | $(x+1)(x^4-x^3+x^2-x-1)$ | $3$ |
Additional Notes
- This same limit can be evaluated with L'Hôpital's Rule (differentiating numerator and denominator once and substituting $x=-1$ into $\frac{3x^2-2}{5x^4-2}$), which gives the same value $\frac{1}{3}$ — the factoring approach here is a purely algebraic alternative that avoids calculus.
- The Factor Theorem — if $p(a)=0$ then $(x-a)$ divides $p(x)$ — is the key fact justifying why both polynomials split off a factor of $(x+1)$ exactly, with no remainder.
- After cancelling the common factor, the simplified rational function is continuous at $x=-1$, since its denominator there ($3$) is nonzero — this is what allows the final substitution step.
Result
The value of the limit is: