Mathematics · Limits and continuity
Edited by: Hitoiki Arcade / Hitoiki Learning team · Published: · Editorial policy
Substituting x = 1 into a fraction can produce 0/0, yet the fraction can have a limit of 2 as x approaches 1. Whether a value exists at the point and where nearby values approach are different questions.
Make three columns on paper: point value, left-hand limit and right-hand limit. We consider real functions near points with available inputs on both sides. Use substitution and factorization to work through three high-school-level examples with this three-column note.
The first exercise asks for the limit of (x² − 1)/(x − 1) as x approaches 1. The original expression is undefined at x = 1. But x² − 1 = (x − 1)(x + 1), so for x ≠ 1 the common nonzero factor cancels and the fraction equals x + 1. Retain the restriction x ≠ 1 after cancellation.
Inputs 0.9 and 0.99 give outputs 1.9 and 1.99. Inputs 1.1 and 1.01 give 2.1 and 2.01. These suggest an approach to 2 from both sides. A few sample values do not establish the conclusion, so check the distance using the expression itself.
For x ≠ 1, the difference between the output and 2 is (x + 1) − 2 = x − 1. The distances are equal: keeping x within 0.01 of 1 keeps the output within 0.01 of 2. Whatever smaller positive tolerance is specified, keeping the input within that same tolerance meets it. This establishes the limit of 2 in this example.
A limit examines nearby inputs excluding the target input itself. The original fraction therefore has limit 2 even though it has no value at x = 1. The expression 0/0 is undefined; it is neither an answer of zero nor a conclusion that the limit does not exist.
The second exercise defines f(x) = 1 for x < 0 and f(x) = 2 for x ≥ 0. Approaching zero through negative inputs always gives 1. Through positive inputs it always gives 2. The left-hand limit is 1 and the right-hand limit is 2. Since they disagree, the two-sided limit does not exist. The point value f(0) = 2 does not remove the disagreement on the left.
The average 3/2 is not a limit either. Outputs on both sides stay 0.5 away from it, so they cannot come within a tolerance such as 0.1. To sketch the function, draw height 1 on the left with an open circle at (0, 1), and height 2 on the right with a filled circle at (0, 2). The filled circle marks the actual point value; it does not determine a two-sided limit on its own.
The third exercise asks when f is continuous at x = a. Here the function can be considered on both sides of a. The value f(a) must be defined, the limit as x approaches a must exist, and that limit must equal f(a). Having a point value alone is insufficient. Having equal one-sided limits alone is also insufficient.
For example, define a function to be 0 whenever x ≠ a, but 1 at x = a. Both one-sided limits are 0, while the point value is 1, so it is not continuous at a. Similarly, keep the first fraction unchanged for x ≠ 1 and assign a new value of 2 at x = 1: this extension is continuous there. Assigning 0 instead disagrees with the limit 2. Changing only the point value while keeping the nearby destination explains why these are separate questions.
For practice, consider (x² − 4)/(x − 2). The original expression excludes x = 2. Factor and cancel for x ≠ 2, find the limit as x approaches 2, then decide which value at x = 2 would make an extension continuous. Away from 2 the expression is x + 2, so the limit is 4 and assigning the point value 4 makes it continuous. Assigning the point value 0 produces the three columns 0, 4, 4, which do not give continuity.
Now work through the three exercises below in order, writing reasons in your columns before opening the answers. If substitution or equality is unclear, revisit the prerequisite “Expressions: give the unknown a role.” The related “Graphs and coordinates: read what each axis means” helps separate position from value. The practice preset selects standard, high-school-level questions on expressions and functions, including other matching questions beyond these three.
TRY & READ
Compare the choices before opening the answer and explanation. Reading an example does not save a test answer or score.
Example 1 · Algebra and functions
Answer: 2
For x ≠ 1, the expression simplifies to x + 1. Its nearby values approach 2. The expression need not be defined at the point for the limit to exist.
Example 2 · Algebra and functions
Answer: Does not exist
The left-hand limit is 1 and the right-hand limit is 2. Since they differ, no two-sided limit exists.
Example 3 · Algebra and functions
Answer: The limit exists and equals f(a)
Continuity requires f(a) to be defined, the limit as x approaches a to exist, and that limit to equal f(a). Differentiability is a stronger condition and is not required. Equal one-sided limits alone are insufficient.
If f(x) = 0 away from a but f(a) = 1, the limit exists and equals 0 while the function is not continuous at a.
Source checked: 2026-10-20
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