Difficulty: Introductory to Intermediate | Prerequisites: Algebra (factoring polynomials, rational expressions), basic trigonometry
This is where Calculus 1 begins. Limits describe what value a function approaches as the input gets close to some number, and continuity tells you whether the function actually arrives there. Nearly every later topic in the course (derivatives, integrals, series) is built on limits, so the mechanics here need to be automatic. If you missed precalculus review, brush up on factoring and trig identities first.
Limits let you work out what a function's output approaches even when you cannot plug the value straight in. When a limit gives you 0/0, you either factor and cancel or use L'Hôpital's rule. Continuity means the function's limit and its actual value at that point agree, and discontinuities in rational functions come from zeros in the denominator.
Limit
The value that f(x) approaches as x gets arbitrarily close to some number c. Written lim (x→c) f(x). In simple terms, it is the answer to "where is this function heading?" even if it never actually gets there.
Indeterminate form
An expression such as 0/0 or ∞/∞ whose value cannot be read off directly. Think of it as a signal that more work is needed, not an answer in itself.
L'Hôpital's rule
If lim (x→c) f(x)/g(x) gives 0/0 or ∞/∞, the limit equals lim (x→c) f'(x)/g'(x), provided that second limit exists. In simple terms, when you are stuck with 0/0, differentiate the top and bottom separately and try again.
Continuity
A function f is continuous at x = c when three things hold: f(c) is defined, lim (x→c) f(x) exists, and the limit equals f(c). Think of it as being able to draw the graph through that point without lifting your pen.
Discontinuity
A point where at least one of the three continuity conditions fails. For rational functions, discontinuities occur wherever the denominator equals zero.
Factoring technique (for limits)
The method of factoring numerator and denominator, cancelling the common factor that causes 0/0, and then substituting. This is usually the first tool to reach for before L'Hôpital's rule.
When direct substitution gives 0/0, factor both numerator and denominator
Cancel the common factor, then substitute the value of x
Example: lim (x→1) (2x² + x − 3)/(x² − x)
Numerator factors as (2x + 3)(x − 1)
Denominator factors as x(x − 1)
Cancel (x − 1): limit becomes (2(1) + 3)/(1) = 5
Answer: 5
Applies when direct substitution gives 0/0 or ∞/∞
Differentiate the numerator and denominator separately (not as a quotient)
Then re-evaluate the limit
Example: lim (x→0) 8x²/(cos x − 1)
Direct substitution gives 0/0
Differentiate top: 16x. Differentiate bottom: −sin x
New limit: lim (x→0) 16x/(−sin x), still 0/0
Apply again: top becomes 16, bottom becomes −cos x
Result: 16/(−1) = −16
Answer: −16
A rational function p(x)/q(x) is discontinuous wherever q(x) = 0
Set the denominator equal to zero and solve
Example: f(x) = 1/(x² − 3x + 2)
Denominator factors as (x − 1)(x − 2)
Discontinuous at x = 1 and x = 2
The distance between these values is |2 − 1| = 1
Answer: 1
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}L'Hôpital's rule. Valid only when the original limit is an indeterminate form (0/0 or ∞/∞). Can be applied repeatedly if the new limit is still indeterminate.
For discontinuities of a rational function f(x) = p(x)/q(x), solve q(x) = 0. Each solution is a point of discontinuity.
For factoring limits, the general workflow is: substitute, confirm 0/0, factor, cancel the offending term, substitute again.
Students often think 0/0 means the limit "does not exist." It does not. 0/0 is an indeterminate form, which means the limit might exist but you need a different method to find it.
Students sometimes apply L'Hôpital's rule when the form is not indeterminate (e.g. when substitution gives 5/3). The rule only applies to 0/0 or ∞/∞. Using it on a determinate form gives a wrong answer.
When factoring, students cancel terms that are not common factors. You can only cancel a factor that appears in both the numerator and the denominator as a multiplicative factor, not as a term added or subtracted.
Students confuse "the function is undefined at x = c" with "the limit does not exist at x = c." The limit can exist perfectly well at a point where the function itself is not defined.
⚠️ Limits by factoring appear on nearly every Calculus 1 exam. You will be expected to recognise 0/0, factor quickly, and simplify.
⚠️ L'Hôpital's rule questions often require two applications in a row. If the first pass still gives 0/0, apply it again.
⚠️ Discontinuity questions test whether you can factor a denominator and solve for its zeros. They sometimes ask for the distance between the discontinuities rather than the values themselves, so read the question carefully.
⚠️ Limits are the foundation for the definition of the derivative. If you are shaky on limits, differentiation will not make sense.
True or false: If direct substitution gives 0/0, the limit does not exist. (False. 0/0 is indeterminate, meaning the limit may still exist.)
Fill in the blank: L'Hôpital's rule says to differentiate the ______ and ______ separately. (numerator, denominator)
True or false: A function can have a limit at a point where it is not defined. (True.)
Fill in the blank: A rational function is discontinuous wherever its ______ equals zero. (denominator)
True or false: You can always apply L'Hôpital's rule to any limit of a fraction. (False. Only to 0/0 or ∞/∞ forms.)
Q: Evaluate lim (x→1) (2x² + x − 3)/(x² − x).
A: Factor the numerator as (2x + 3)(x − 1) and the denominator as x(x − 1). Cancel (x − 1). The limit is (2 + 3)/1 = 5.
Q: Evaluate lim (x→0) 8x²/(cos x − 1).
A: Direct substitution gives 0/0. Apply L'Hôpital's rule: 16x/(−sin x), still 0/0. Apply again: 16/(−cos x). At x = 0 this gives 16/(−1) = −16.
Q: Find the distance between the two discontinuities of f(x) = 1/(x² − 3x + 2).
A: Factor the denominator: (x − 1)(x − 2). Discontinuities at x = 1 and x = 2. Distance = |2 − 1| = 1.
Q: Suppose lim (x→c) f(x) = 4 but f(c) is undefined. Is f continuous at c?
A: No. Continuity requires f(c) to be defined and equal to the limit. Since f(c) is undefined, the function is discontinuous at c.
Limit evaluation, indeterminate form, 0/0 limit, L'Hôpital's rule, L'Hopital's rule, factoring limits, cancelling common factors, continuity at a point, discontinuity of rational functions, zeros of the denominator, removable discontinuity, Calculus 1 limits review, MATH 104 Penn State, finding limits algebraically