Does a piecewise limit exist at x = 1?
Let $$ f(x) = \begin{cases} x^2 + 1 & \text{for } x < 1, \\[4pt] \dfrac{x^2 - 1}{x - 1} & \text{for } x > 1. \end{cases} $$ Determine whether $\displaystyle\lim_{x \to 1} f(x)$ exists, and find its value.
Explanation of Method
A two-sided limit exists at a point exactly when the one-sided limits from the left and from the right both exist and agree. Since $f$ is defined by two different formulas on either side of $x = 1$, the natural approach is to compute the left-hand limit using the first formula and the right-hand limit using the second, then compare the two results.
- Evaluate the left-hand limit using the polynomial piece $x^2+1$.
- Simplify the rational piece $\dfrac{x^2-1}{x-1}$ by factoring, then evaluate the right-hand limit.
- Compare the two one-sided limits to determine whether the two-sided limit exists.
Step 1 · Set up the comparison
The function is not defined at $x = 1$ itself, so to determine the behavior of $f$ there we check whether:
$$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x). $$
If the two one-sided limits agree, that common value is $\displaystyle\lim_{x\to 1} f(x)$.
Step 2 · Compute the left-hand limit
For $x < 1$, $f(x) = x^2 + 1$, which is a polynomial and hence continuous everywhere, so the limit is found by direct substitution:
$$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 1) = 1^2 + 1 = 2. $$
Step 3 · Compute the right-hand limit
For $x > 1$, $f(x) = \dfrac{x^2-1}{x-1}$. Since $x \neq 1$ on this side, the numerator can be factored and the common factor cancelled:
$$ x^2 - 1 = (x-1)(x+1) \quad\Longrightarrow\quad f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1 \quad \text{for } x > 1. $$
Now the limit is found by direct substitution in the simplified expression:
$$ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x+1) = 1 + 1 = 2. $$
Step 4 · Compare the one-sided limits
Both one-sided limits agree:
$$ \lim_{x \to 1^-} f(x) = 2 \qquad \text{and} \qquad \lim_{x \to 1^+} f(x) = 2. $$
Since the left-hand and right-hand limits are equal, the two-sided limit exists and equals their common value. (Note that this also means $f$ could be extended to a function that is continuous at $x=1$ by defining $f(1) = 2$, even though $f$ is not defined there as given.)
Summary Table
| Side | Formula used | Limit as $x \to 1$ |
|---|---|---|
| Left ($x<1$) | $x^2+1$ | $2$ |
| Right ($x>1$) | $x+1$ (after cancelling $x-1$) | $2$ |
Additional Notes
- The right-hand piece $\dfrac{x^2-1}{x-1}$ is undefined at $x=1$, but factoring reveals a removable discontinuity: the simplified form $x+1$ has the same limit there.
- Checking both one-sided limits is essential whenever a function is defined piecewise across a boundary point — agreement of the two sides is both necessary and sufficient for the two-sided limit to exist.
- This problem illustrates the difference between the limit of $f$ at a point and the value of $f$ at that point: the limit can exist even when $f(1)$ is not defined.
Result
The limit exists and its value is: