Calculus Honors · AP-aligned study guide

1.1 · Understanding limits: graphs, tables, and one-sided behavior

Describe what a function approaches, distinguish a limit from a function value, and connect graphical and numerical evidence.

AP topics 1.1–1.4 and 1.9 · Reading: Larson & Edwards §§1.1–1.2; one-sided limit ideas from §1.4

Learning objectives

Read limit notation and distinguish approach from evaluation.
Estimate limits using values on both sides of a point.
Use matching one-sided limits to decide whether a two-sided limit exists.

Why calculus begins with approach

Average velocity measures change over a time interval. Instantaneous velocity asks what those averages approach as the interval shrinks toward zero. We cannot divide by an interval of zero; instead, we study a limit. The same idea turns secant slopes into tangent slopes in the next unit.

The statement $\displaystyle \lim_{x\to a}f(x)=L$ says that f(x) can be made as close as desired to L by choosing x sufficiently close to a, with $\displaystyle x\ne a$ . Here a is an input and L is an output. The function need not be defined at a for its limit to exist.

$s'(t)=\frac{ds}{dt}=\lim_{h\to0}\frac{s(t+h)-s(t)}{h}$

A hole does not determine the limit

For f(x) = $\displaystyle \frac{x^2-4}{x-2}$ , the original expression is undefined at x = 2. For every nearby $\displaystyle x\ne2$ , factoring gives f(x) = x + 2. The outputs therefore approach 4. The graph is the line y = x + 2 with an open circle at (2, 4).

If we define a new function g that agrees with f away from 2 but has g(2) = 1, the limit is still 4. A filled point records g(2); the surrounding curve records the limit. Always answer the quantity the question asks for.

$\lim_{x\to2}\frac{x^2-4}{x-2}=4$

For g, the limit is 4 while g(2) = 1.

Build a table from both directions

Use inputs approaching 2 from below and above. In the table, both sets of outputs move toward 4. This is evidence for a limit, while the simplified expression x + 2 supplies the algebraic justification.

A finite table alone cannot prove a limit. A function may behave differently between sampled inputs. Avoid entering the exact input when the expression is undefined, and keep enough digits to see the trend.

Approaching x = 2 from both sides
x	1.9	1.99	1.999	2.001	2.01	2.1
f(x)	3.9	3.99	3.999	4.001	4.01	4.1

One-sided limits must agree

The notation $\displaystyle x\to a^-$ means approach a using smaller x-values. The notation $\displaystyle x\to a^+$ means approach using larger x-values. The superscript tells you the direction of the input; it does not tell you whether the output is positive or negative.

A finite two-sided limit exists exactly when both one-sided limits exist and equal the same finite number. If a graph approaches height 2 from the left and height 5 from the right, the two-sided limit does not exist. Choosing a value at the point cannot repair that disagreement.

$\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L$

Write a complete AP-style conclusion

State each one-sided value before concluding whether the two-sided limit exists. For a graph, trace the curve toward the input from each side; do not follow a filled dot away from the surrounding curve. At an endpoint of a domain, identify the available one-sided limit explicitly.

When the outputs grow without bound, there is no finite limit. Infinite-limit notation describes unbounded behavior; infinity is not a real-number height. We will study this and vertical asymptotes in a later Unit 1 lesson.

Worked examples

A removable hole with a different point value

Let $\displaystyle g(x)=x+2$ for $\displaystyle x\ne2$ , and $\displaystyle g(2)=1$ . Find both one-sided limits, the two-sided limit, and $\displaystyle g(2)$ .

From the left, $\displaystyle x+2$ approaches $\displaystyle 4$ : $\displaystyle \lim_{x\to2^-}g(x)=4$ .
From the right, $\displaystyle x+2$ also approaches $\displaystyle 4$ : $\displaystyle \lim_{x\to2^+}g(x)=4$ .
The one-sided limits agree, so $\displaystyle \lim_{x\to2}g(x)=4$ .
The definition gives $\displaystyle g(2)=1$ . The point value does not change the limit.

A jump in a piecewise function

Let $\displaystyle h(x)=x+1$ for $\displaystyle x<1$ , and $\displaystyle h(x)=4-x$ for $\displaystyle x\geq1$ . Does $\displaystyle \lim_{x\to1}h(x)$ exist?

Use the left branch: $\displaystyle \lim_{x\to1^-}h(x)=2$ .
Use the right branch: $\displaystyle \lim_{x\to1^+}h(x)=3$ .
Since $\displaystyle 2\ne3$ , the two-sided limit does not exist.
Although $\displaystyle h(1)=3$ , the point value cannot settle the two-sided limit.

Common mistakes

Reading a filled dot as the limit instead of tracing nearby behavior.
Using only one side of a table to claim a two-sided limit.
Saying a limit does not exist merely because the function is undefined at the point.

Try it yourself

No calculator required.

1. For $\displaystyle f(x)=\frac{x^2-9}{x-3}$ , $\displaystyle x\ne3$ , make a table at $\displaystyle x=2.9,\ 2.99,\ 3.01,\ 3.1$ . Estimate $\displaystyle \lim_{x\to3}f(x)$ .

Show worked solution

The outputs are $\displaystyle 5.9,\ 5.99,\ 6.01,\ 6.1$ . Both sides suggest $\displaystyle 6$ . Factoring gives $\displaystyle f(x)=x+3$ for $\displaystyle x\ne3$ , so $\displaystyle \lim_{x\to3}f(x)=6$ .

2. Use the graph of $\displaystyle f$ below. It follows $\displaystyle y=2x-1$ , with an open circle at $\displaystyle (2,3)$ and a filled dot at $\displaystyle (2,7)$ . Find $\displaystyle \lim_{x\to2^-}f(x)$ , $\displaystyle \lim_{x\to2^+}f(x)$ , $\displaystyle \lim_{x\to2}f(x)$ , and $\displaystyle f(2)$ .

Line y = 2x minus 1 with an open circle at (2, 3) and a filled point at (2, 7). — Trace the line toward x = 2 from both directions.

Show worked solution

Tracing the line toward $\displaystyle x=2$ from either side gives $\displaystyle \lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=3$ . Therefore $\displaystyle \lim_{x\to2}f(x)=3$ . The filled dot gives $\displaystyle f(2)=7$ ; its height does not change the limit.

Line y = 2x minus 1 with an open circle at (2, 3) and a filled point at (2, 7). Both sides approach height 3; the filled point gives f(2) = 7. — Both sides approach height 3; the filled point gives f(2) = 7.

3. Let $\displaystyle p(x)=2x$ for $\displaystyle x<1$ and $\displaystyle p(x)=x+3$ for $\displaystyle x\geq1$ . Use the graph and the formulas to find both one-sided limits at $\displaystyle 1$ . Does $\displaystyle \lim_{x\to1}p(x)$ exist?

Left branch y = 2x ends with an open circle at (1, 2); right branch y = x plus 3 starts with a filled point at (1, 4). — Open and filled endpoints identify which branch includes the join.

Show worked solution

The left branch approaches $\displaystyle 2$ , while the right branch approaches $\displaystyle 4$ . Thus $\displaystyle \lim_{x\to1^-}p(x)=2$ and $\displaystyle \lim_{x\to1^+}p(x)=4$ . Since these differ, the two-sided limit does not exist, even though $\displaystyle p(1)=4$ .

Left branch y = 2x ends with an open circle at (1, 2); right branch y = x plus 3 starts with a filled point at (1, 4). The left and right approached heights differ, so there is no two-sided limit. — The left and right approached heights differ, so there is no two-sided limit.

4. For $\displaystyle s(t)=t^2+2t$ , the average velocity on $\displaystyle [1,1+h]$ is $\displaystyle \frac{s(1+h)-s(1)}{h}=4+h$ for $\displaystyle h\ne0$ . Evaluate it for $\displaystyle h=0.1,\ 0.01,\ -0.01$ . Find the instantaneous velocity $\displaystyle s'(1)$ .

Show worked solution

The averages are $\displaystyle 4.1,\ 4.01,\ 3.99$ . The derivative is the limit of these average rates: $\displaystyle s'(1)=\lim_{h\to0}\frac{s(1+h)-s(1)}{h}=\lim_{h\to0}(4+h)=4$ . The tangent line at $\displaystyle (1,3)$ has slope $\displaystyle 4$ , in distance units per time unit.

Parabola s(t) = t squared plus 2t with a tangent line at (1, 3). The tangent has rise 4 over run 1, matching the derivative s prime of 1 = 4. — The tangent has rise 4 over run 1, matching the derivative s prime of 1 = 4.

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