Limit of (x² − 1)/(2x² − x − 1) at 0
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{x^2 - 1}{2x^2 - x - 1}. $$
Explanation of Method
The first thing to check in any rational-function limit is whether the denominator vanishes at the target point. Here the denominator at $x=0$ equals $-1 \neq 0$, so the function is already continuous there and the limit can be found by direct substitution — no factoring is actually required. Factoring both polynomials is nonetheless instructive, since it reveals the full simplified form of the function and shows exactly where the genuine discontinuity (at $x=1$) lies.
- Check the denominator at $x=0$ to confirm direct substitution is valid.
- Factor the numerator and denominator to obtain the simplified form of the function.
- Evaluate the limit either by direct substitution or from the simplified form.
Step 1 · Check the denominator at $x=0$
Substituting $x=0$ into the denominator:
$$ 2(0)^2 - 0 - 1 = -1 \neq 0. $$
Since the denominator does not vanish at $x=0$, the rational function $\dfrac{x^2-1}{2x^2-x-1}$ is continuous at $x=0$, and the limit equals the value of the function there:
$$ \lim_{x\to0}\frac{x^2-1}{2x^2-x-1} = \frac{0^2-1}{2(0)^2-0-1} = \frac{-1}{-1} = 1. $$
Step 2 · Factor both polynomials (for insight)
The numerator is a difference of squares:
$$ x^2 - 1 = (x-1)(x+1). $$
The denominator factors by finding two numbers that multiply to $2\times(-1)=-2$ and add to $-1$, namely $-2$ and $1$:
$$ 2x^2 - x - 1 = (2x+1)(x-1). $$
So the function simplifies, for $x \neq 1$, to:
$$ \frac{x^2-1}{2x^2-x-1} = \frac{(x-1)(x+1)}{(2x+1)(x-1)} = \frac{x+1}{2x+1}. $$
Step 3 · Confirm the limit from the simplified form
Since $x=0$ is not the excluded point $x=1$, the simplified expression is valid there and gives the same result:
$$ \lim_{x \to 0} \frac{x+1}{2x+1} = \frac{0+1}{2(0)+1} = \frac{1}{1} = 1. $$
Summary Table
| Approach | Expression used at $x=0$ | Result |
|---|---|---|
| Direct substitution | $\dfrac{x^2-1}{2x^2-x-1}$ | $1$ |
| Factor & simplify | $\dfrac{x+1}{2x+1}$ | $1$ |
Additional Notes
- Whenever a rational function's denominator is nonzero at the target point, the limit is simply the value of the function there — factoring is unnecessary, though it never hurts to confirm.
- The factored form $\dfrac{x+1}{2x+1}$ shows the original function actually has a removable discontinuity at $x=1$ (where both the original numerator and denominator vanish), not at $x=0$; this problem's limit point is unrelated to that discontinuity.
- Factoring is essential only when direct substitution produces $\frac{0}{0}$ or another indeterminate form — always check the denominator first before reaching for algebraic manipulation.
Result
The value of the limit is: