In Limits from Both Sides we let $x$ approach $a$ from both sides at once. Sometimes the two sides tell different stories, so we give each side its own notation.
🔑 Key idea
$\displaystyle\lim_{x \to a^-} f(x)$ uses only $x < a$ (from the left); $\displaystyle\lim_{x \to a^+} f(x)$ uses only $x > a$ (from the right). The two-sided limit $\lim_{x \to a} f(x) = L$ exists exactly when both one-sided limits exist and equal $L$.
Think of two hikers walking toward the same spot on a trail, one from the west and one from the east. The two-sided limit asks whether they arrive at the same height. If the trail has a cliff at that spot, each hiker still ends somewhere definite, but they end at different heights.
A jump at $x = 1$: from the left the values approach $2$, from the right they approach $3$. The filled value $f(1) = 3$ belongs to the right-hand piece.
💡 Why it works
The definition of a limit asks that $f(x)$ be close to a single number $L$ for all $x$ near $a$, on both sides. If the left side settles at 2 and the right side at 3, then no single $L$ is close to both, so the two-sided limit cannot exist. When both sides agree, every nearby $x$ gives a value near that common number, and the two-sided limit exists.
✏️ Worked example: A piecewise function
Let $$f(x) = \begin{cases} x^2 - 1 & x < 2 \\ 2x + 1 & x \ge 2 \end{cases}$$. Find the left, right and two-sided limits at $x = 2$.
From the left ($x < 2$) use $x^2 - 1$: it tends to $4 - 1 = 3$.
From the right ($x > 2$) use $2x + 1$: it tends to $4 + 1 = 5$.
The sides disagree ($3 \ne 5$), so the graph jumps at $x = 2$.
Answer: $\lim_{x \to 2^-} f(x) = 3$, $\lim_{x \to 2^+} f(x) = 5$, and $\lim_{x \to 2} f(x)$ does not exist.
✏️ Worked example: Absolute value on the bottom
Find the one-sided limits of $f(x) = \dfrac{x^2 - 9}{\lvert x - 3 \rvert}$ at $x = 3$.
Factor the top: $x^2 - 9 = (x - 3)(x + 3)$.
For $x > 3$, $\lvert x - 3 \rvert = x - 3$, so $f(x) = x + 3 \to 6$.
For $x < 3$, $\lvert x - 3 \rvert = -(x - 3)$, so $f(x) = -(x + 3) \to -6$.
Different one-sided limits, so no two-sided limit.
Answer: Left limit $-6$, right limit $6$; the two-sided limit does not exist.
✏️ Worked example: When the sides agree
Let $f(x) = 3x - 1$ for $x < 1$ and $f(x) = x^2 + 1$ for $x > 1$, with $f(1)$ undefined. Find $\lim_{x \to 1} f(x)$.
Left: $3(1) - 1 = 2$. Right: $1^2 + 1 = 2$.
Both sides give 2, so the two-sided limit exists, even though $f(1)$ is not defined.
Answer: $\lim_{x \to 1} f(x) = 2$
⚠️ Common mistake
At a jump, a common slip is to compute one side, usually by plugging $a$ into whichever formula has the "$\ge$" sign, and report it as the two-sided limit. Always compute both sides. A quick check: if the formula changes at $x = a$, the question "which piece do I use?" has two answers, so you need two one-sided limits.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 1 XP.