Simplify a fraction and take its limit at 1
Consider the function: $$ f(x) = \frac{x(x^2-1)}{(x^2+1)(x-1)}. $$ Simplify $f(x)$ and determine $\displaystyle\lim_{x \to 1} f(x)$.
Explanation of Method
The numerator contains a hidden difference of squares, $x^2-1$, which shares a factor of $(x-1)$ with the denominator. Factoring and cancelling that common factor produces a simplified function that agrees with $f$ everywhere except at $x=1$, where the original $f$ is undefined (division by zero) but the simplified form is perfectly continuous. The limit at $x=1$ is then read off directly from the simplified, continuous expression.
- Factor the numerator using the difference of squares.
- Cancel the common factor $(x-1)$ shared with the denominator.
- Evaluate the limit at $x=1$ using the simplified, continuous expression.
Step 1 · Factor the numerator
The numerator is $x(x^2-1)$. Since $x^2-1$ is a difference of squares:
$$ x^2 - 1 = (x-1)(x+1) \quad\Longrightarrow\quad x(x^2-1) = x(x-1)(x+1). $$
The denominator is already factored: $(x^2+1)(x-1)$. So:
$$ f(x) = \frac{x(x-1)(x+1)}{(x^2+1)(x-1)}. $$
Step 2 · Cancel the common factor
For $x \neq 1$, the factor $(x-1)$ appears in both the numerator and denominator and can be cancelled:
$$ f(x) = \frac{x(x+1)}{x^2+1}, \qquad \text{for } x \neq 1. $$
The original $f$ is undefined at $x=1$ (both numerator and denominator of the un-simplified form vanish there — a $\frac00$ situation), but the simplified expression $\dfrac{x(x+1)}{x^2+1}$ is a polynomial ratio with denominator $x^2+1 > 0$ for all real $x$, so it is defined and continuous everywhere, including at $x=1$.
Step 3 · Evaluate the limit at $x=1$
Since the simplified expression is continuous at $x=1$, the limit is found by direct substitution:
$$ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x(x+1)}{x^2+1} = \frac{1(1+1)}{1^2+1} = \frac{2}{2} = 1. $$
Summary Table
| Form | Expression | Defined at $x=1$? |
|---|---|---|
| Original | $\dfrac{x(x^2-1)}{(x^2+1)(x-1)}$ | No ($\frac00$) |
| Simplified | $\dfrac{x(x+1)}{x^2+1}$ | Yes, value $=1$ |
Additional Notes
- The point $x=1$ is a removable discontinuity of the original $f$: the function can be "repaired" by defining $f(1)=1$, after which it agrees exactly with the simplified expression and becomes continuous everywhere.
- Note that $x^2+1$ never equals zero for real $x$, so the simplified function $\dfrac{x(x+1)}{x^2+1}$ has no discontinuities at all — the only irregularity of the original $f$ came from the factor $(x-1)$ that was common to both numerator and denominator.
- This is a canonical example of the general principle: whenever a rational function's numerator and denominator share a common linear factor, that factor produces a removable discontinuity rather than a vertical asymptote, and the limit there equals the value of the simplified function.
Result
The simplified function and the limit are: