Connecting Multiple Representations of Limits

A limit question on the exam can hand you a formula, a graph, a table of values, or a sentence such as "as $x$ approaches 2, $f(x)$ approaches 3." These are four views of one object, and a question often gives you two of them at once, say $f$ as a graph and $g$ as a formula, and asks about $f(g(x))$.

One limit, four views

The statement $\lim\limits_{x \to c} f(x) = L$ says that the outputs $f(x)$ get as close to $L$ as you want once the inputs $x$ are close enough to $c$, with $x \neq c$. The value $f(c)$ plays no part: it can equal $L$, differ from $L$, or not exist. The one-sided limits $\lim\limits_{x \to c^-} f(x)$ and $\lim\limits_{x \to c^+} f(x)$ use only inputs below $c$ or only inputs above $c$, and the two-sided limit exists exactly when both one-sided limits exist and are equal.

Take $h(x) = \dfrac{x^3 - 8}{x^2 - 4}$ near $x = 2$, where it is undefined because the denominator is 0.

Each row says the same thing. Translating between them is mechanical once you know what to look for: a limit on a graph is a height the curve heads toward, in a table it is the value the outputs settle on from both directions, in a formula it is what the algebra of topics 1.5 and 1.6 produces, and in words it is the sentence the notation abbreviates.

What each view can prove

The views are not equally strong. A formula, worked with the limit laws, algebra, or the squeeze theorem (topic 1.8), proves a limit's value. A graph whose key heights are labeled or sit on grid lines, or a sentence stating a limit, hands you an exact value to use as given, the way a theorem's hypothesis is given. A table, alone, only suggests a value.

A table samples, it doesn't prove. A table shows finitely many inputs, and a function can do anything between them. Let $q(x) = \cos\left(\dfrac{\pi}{x}\right)$. At $x = \pm 0.1, \pm 0.01, \pm 0.001$ the inputs to the cosine are $\pm 10\pi, \pm 100\pi, \pm 1000\pi$, all even multiples of $\pi$, so every entry of the table is exactly 1. Yet $\lim\limits_{x \to 0} q(x)$ does not exist: at $x = \dfrac{1}{3}, \dfrac{1}{5}, \dfrac{1}{7}, \ldots$, which also crowd toward 0, the cosine's input is an odd multiple of $\pi$ and $q(x) = -1$. The table chose inputs that hid the oscillation. That is why exam questions ask you to estimate a limit from a table, and why a table is evidence you weigh against other information (for instance, being told the function is continuous) rather than a proof.

A graph is read, not computed. On a graph the limit from the left is the height the curve approaches as you trace it toward $x = c$ from the left, and likewise from the right. An open circle marks a point missing from the graph, and a filled dot marks the actual value $f(c)$. The limit ignores both, because it is about the heights near $c$. A drawing is only as precise as its labels: take heights from labeled points and grid lines, and when a figure is marked not drawn to scale, trust the labels over the picture. A limit read that way is exact, which is why Example 2 below combines graph-read limits with the limit laws.

Infinite behavior in every view. If the outputs grow without bound as $x \to c^+$, you write $\lim\limits_{x \to c^+} f(x) = \infty$. On a graph this is a curve climbing along a vertical asymptote; in a table, outputs like $40, 400, 4000$ as $x$ gets closer to $c$; in a formula, typically a nonzero numerator over a denominator approaching 0. Writing $\infty$ describes how the limit fails to exist; it does not make the limit a number. Topic 1.14 covers infinite limits in full.

Combining limits from different views

The limit laws do not care where a limit came from. If $\lim\limits_{x \to c} f(x) = 4$ is read off a graph and $\lim\limits_{x \to c} g(x) = -1$ comes from a formula, then $\lim\limits_{x \to c}[f(x)g(x)] = -4$. The laws need each piece's limit to exist, though, and when one doesn't, the combination can still have a limit. You find it by going back to one-sided limits, combining each side separately, and comparing.

Compositions need the direction of approach. To find $\lim\limits_{x \to c} f(g(x))$, first find what $g(x)$ approaches and from which side, then ask what $f$ does as its input approaches that value from that side. With the graph of $f$ in Example 2 below, which jumps at 3 (heights near 4 just left of 3, heights near 3 just right of 3), compare two limits at $x = 0$:

• In $f(3 - x^2)$, the input $3 - x^2$ approaches 3 and is always less than 3 for $x \neq 0$, so $\lim\limits_{x \to 0} f(3 - x^2) = \lim\limits_{u \to 3^-} f(u) = 4$.
• In $f(3 + x^2)$, the input approaches 3 from above, so $\lim\limits_{x \to 0} f(3 + x^2) = \lim\limits_{u \to 3^+} f(u) = 3$.

The inside approaches the same number in both, and the answers differ only because of the side. If the inside approaches a value $L$ while never equaling it, $f(L)$ itself is irrelevant; you need the limit of $f$ at $L$, not its value.

Worked examples

Worked exampleExample 2: reading a graph, and combining it with a formula

The graph of $f$ on $0 \le x \le 6$ is shown below. Let $g(x) = \dfrac{|x - 3|}{2(x-3)}$.

(a) Find $\lim\limits_{x \to 1} f(x)$ and $f(1)$.

(b) Find $\lim\limits_{x \to 3^-} f(x)$, $\lim\limits_{x \to 3^+} f(x)$, and $\lim\limits_{x \to 3} f(x)$ if it exists.

(c) Find $\lim\limits_{x \to 3}\left[f(x) + g(x)\right]$.

Show solution

(a) As $x$ approaches 1 from either side, the curve's height approaches the open circle at 2, so $\lim\limits_{x \to 1} f(x) = 2$. The filled dot gives $f(1) = 4$. The two numbers differ, and that is allowed: the limit describes the heights near 1, the value is the height at 1.

(b) Tracing the left segment toward $x = 3$, the heights approach 4, so $\lim\limits_{x \to 3^-} f(x) = 4$. Tracing the right segment toward 3, they approach 3, so $\lim\limits_{x \to 3^+} f(x) = 3$. The one-sided limits are different, so $\lim\limits_{x \to 3} f(x)$ does not exist.

(c) Neither $f$ nor $g$ has a limit at 3, so the sum law doesn't apply; work each side. For $x < 3$, $|x - 3| = -(x - 3)$, so $g(x) = -\frac{1}{2}$; for $x > 3$, $g(x) = \frac{1}{2}$. Then

$$\lim_{x \to 3^-}\left[f(x) + g(x)\right] = 4 - \tfrac{1}{2} = \tfrac{7}{2}, \qquad \lim_{x \to 3^+}\left[f(x) + g(x)\right] = 3 + \tfrac{1}{2} = \tfrac{7}{2}.$$

The one-sided limits agree, so $\lim\limits_{x \to 3}\left[f(x) + g(x)\right] = \dfrac{7}{2}$. The jump of $-1$ in $f$ is cancelled by the jump of $+1$ in $g$.

Where students lose points

On the exam

A question on this topic gives you a graph or a table of $f$, sometimes with a formula for another function, and asks about a sum, product, quotient or composition. Answer choices can include the function's value at the point and "does not exist," so decide on a two-sided limit only after you have both one-sided limits. When a question gives a table and says the function is continuous at a tabulated input $c$, continuity makes the entry $f(c)$ equal to $\lim\limits_{x \to c} f(x)$ (topic 1.11); the entries at other inputs near $c$ still only suggest a limit.