Limits & Continuity — Print
Where a function is heading matters more than whether it ever arrives.
What the practice questions cover
These questions use 8 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.
- Mean — Two kinds of centre: the balance point, and the middle value.
- Factor — A number that divides evenly into another.
- Function — A rule pairing each input with exactly one output.
- Polynomial — A sum of terms, each a number times a whole-number power of the variable.
- Limit — The value a function approaches as the input approaches some number.
- One-sided limit — The value approached from the left, or from the right, alone.
- Continuous — A function you could draw without lifting the pen — no holes, jumps or blow-ups.
- Indeterminate form — 00, which means the limit needs more work — not that it fails to exist.
Try x = 1 in x2-1x-1.
The bottom becomes zero, so the expression has no value there. The calculator refuses.
Now creep up on it. At 0.9 you get 1.9; at 0.99, 1.99; at 0.999, 1.999.
Come down from above and the same thing happens. Both sides are heading for exactly 2.
The function never reaches 2 at x=1 — it is not defined there at all. But it is unmistakably going there, and a limit is the word for that.
What a limit is asking
limx → a f(x) = L means: as x gets close to a, f(x) gets close to L.
It says nothing about f(a) itself. The function may be undefined there, or defined to be something else entirely.
That separation is the whole point — it lets you talk about a value the function never takes.
Near, not at
The limit ignores the single point x=a on purpose. A hole in the graph does not change where the curve is heading.
Just substitute, first
For most functions at most points, the limit is simply f(a). Substitute and you are done.
This works for every polynomial, and for rationals wherever the bottom is not zero.
When substitution fails, inspect the expression: an indeterminate form, a jump or unbounded behaviour may require another method.
Try the easy thing
Substitution answers most limit questions in one line. Reach for algebra only when it fails.
When you get 0/0
00 is not an answer. It is an indeterminate form. In the rational examples here, factoring may reveal a common factor to cancel; other functions can need different methods.
Factor both, cancel the common piece, and substitute into what is left.
x2-1x-1 = (x-1)(x+1)x-1 = x+1 away from x=1, and x+1 is 2 there.
It is a signal, not a result
Writing "the limit is 00" says only that you have not finished. 00 can turn out to be any number at all, depending on the functions.
One-sided limits, and when the limit fails
Approach from the left and from the right separately. If the two disagree, the limit does not exist.
A step function jumps: coming from the left gives one height, from the right another.
For the limit to exist, both sides must head for the same place.
Three ways it can fail
The sides disagree, the function blows up to infinity, or it oscillates without settling. Anything else, and the limit exists.
Continuity
A function is continuous at a when three things hold: f(a) exists, the limit exists, and they are equal.
A hole fails the first. A jump fails the second. A misplaced single point fails the third.
Continuous functions are the well-behaved ones, and almost every rule later in calculus assumes continuity.
The pen test
Could you draw it without lifting the pen? That is continuity, in a sentence — and it is exactly the three conditions above.
Worked examples
Substitution works
Find limx → 3 (x2 - 2x + 4).
- This is a polynomial, so it is continuous everywhere. Substitution is valid.
- 32 - 2(3) + 4 = 9 - 6 + 4. Just evaluate.
- = 7. No algebra needed.
7
A 0/0 form
Find limx → 2 x2-4x-2.
- Substituting gives 00, so there is a common factor. A signal to factor.
- (x-2)(x+2)x-2 = x+2 for every x ≠ 2. Cancel the shared factor.
- Now substitute: 2+2 = 4. The function has a hole at x=2, and still heads for 4.
4
A limit that does not exist
For f(x) = 2x when x<1 and f(x) = x+5 when x ≥ 1, find limx → 1 f(x).
- From the left, f follows 2x, heading for 2. Left-hand limit.
- From the right, f follows x+5, heading for 6. Right-hand limit.
- The two disagree, so the limit does not exist. The graph jumps at x=1.
Does not exist
Words to know
- Limit
- The value a function approaches as the input approaches some number.
- One-sided limit
- The value approached from the left, or from the right, alone.
- Continuous
- A function you could draw without lifting the pen — no holes, jumps or blow-ups.
- Removable discontinuity
- A single missing point; a hole that could be filled in.
- Indeterminate form
- 00, which means the limit needs more work — not that it fails to exist.
Mistakes to watch for
Reporting 00 as the answer. It is an indeterminate form, meaning the work is unfinished. Factoring and cancelling turns x2-4x-2 at x=2 into a clean 4.
Assuming the limit is f(a). For x2-1x-1 at x=1 the function has no value, yet the limit is 2. The limit is about the approach, not the arrival.
Checking only one side. A step function heads for 2 from the left and 6 from the right, so no limit exists — one side alone would have given a confident wrong answer.
Cancelling and forgetting the hole. x2-4x-2 equals x+2 everywhere except x=2, where the original is still undefined. The limit exists; the value does not.
Confusing a limit with continuity. A limit can exist where the function is undefined. Continuity needs the value to exist and match.
Newton and Leibniz built calculus on this idea in the 1600s, and could not make it rigorous. It took until the 1820s and Cauchy for limits to be defined properly.
Bishop Berkeley mocked the early version, calling infinitesimals "the ghosts of departed quantities". He was right, and it took two centuries to answer him.
The hole at x=1 is real but invisible on a graphing calculator, whose pixels are wider than a single point.
Problem 1
Find limx → 3(2x+1).
Worked answer
7
2(3)+1 = 7
Problem 2
Find limx → 2(x2-5).
Worked answer
-1
4-5 = -1
Problem 3
Find limx → 0(3x2+4x+9).
Worked answer
9
Only the constant survives
Problem 4
Find limx → 4x+2x-1.
Worked answer
2
63 = 2
Problem 5
What does limx → a f(x) = L tell you about f(a)?
- Nothing
- That f(a)=L
- That f(a) is undefined
- That f(a) ≠ L
Worked answer
Nothing
The single point x=a is deliberately ignored
Problem 6
Find limx → 2x2-4x-2.
Worked answer
4
(x-2)(x+2)x-2 = x+2
At x=2 that is 4
Problem 7
Find limx → 3x2-9x-3.
Worked answer
6
(x-3)(x+3)x-3 = x+3
At x=3 that is 6
Problem 8
Find limx → 1x2-1x-1.
Worked answer
2
x+1 away from x=1
At x=1 that is 2
Problem 9
Find limx → -2x2+5x+6x+2.
Worked answer
1
(x+2)(x+3)x+2 = x+3
At x=-2 that is 1
Problem 10
Substituting gives 00. What does that mean?
- More work is needed
- The limit is 0
- The limit is 1
- The limit does not exist
Worked answer
More work is needed
It is an indeterminate form
Top and bottom share a factor that has not been cancelled
Problem 11
For f(x)=2x when x<1 and f(x)=x+5 when x ≥ 1, find the limit as x approaches 1 from the left.
Worked answer
2
2(1) = 2
Problem 12
For the same function, find the limit as x approaches 1 from the right.
Worked answer
6
1+5 = 6
Problem 13
For that same function, does limx → 1 f(x) exist?
- No — the two sides disagree
- Yes, it is 2
- Yes, it is 6
- Yes, it is 4
Worked answer
No — the two sides disagree
2 from the left and 6 from the right
They must agree for the limit to exist
Problem 14
A function has a hole at x=2 but heads for 5 from both sides. Which is true?
- The limit is 5 but it is not continuous there
- It is continuous
- The limit does not exist
- The limit is 0
Worked answer
The limit is 5 but it is not continuous there
The limit exists; the value does not
Problem 15
Find limx → 0x2+3xx.
Worked answer
3
x(x+3)x = x+3
At x=0 that is 3
Problem 16
Find limx → 4x-4√x-2.
Worked answer
4
Bottom becomes x-4
So the expression is √x + 2, which is 4 at x=4