Limits & Continuity — Challenge
Where a function is heading matters more than whether it ever arrives.
What these problems 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.
Problem 1
What does x2-4x-2 approach as x approaches 2?
- 0
- 2
- 4
- 8
- It is undefined
Worked answer
4
Substituting x=2 directly gives 00, which tells us nothing on its own.
Factorise instead: (x-2)(x+2)x-2 = x+2 for every x other than 2.
As x approaches 2, that expression approaches 4. The function has no value at x=2, but it has a limit there.
Problem 2
Find limx → 0 √x+9-3x.
- 16
- 0
- 13
- 6
- It does not exist
Worked answer
16
Substituting gives 00, so the expression needs rearranging. Multiply top and bottom by √x+9+3. The top becomes (x+9)-9 = x. So the expression is 1√x+9+3, which at x=0 is 13+3 = 16.
Problem 3
Find limx → ∞ 3x2+1x2-5.
- 3
- 0
- ∞
- 13
- -15
Worked answer
3
Far from the origin the highest powers dominate everything else. Divide top and bottom by x2: 3 + 1/x21 - 5/x2. As x grows the small terms vanish, leaving 31 = 3.
Problem 4
Find limx → 3 x-3x2-9.
- 16
- 0
- 6
- 13
- It does not exist
Worked answer
16
Substituting gives 00, so factor the bottom. x-3(x-3)(x+3) = 1x+3 for every x ≠ 3. At x=3 that is 16.
Problem 5
Let f(x) = x2-4x-2 for x ≠ 2, and f(2) = k. For which k is f continuous at 2?
- 4
- 2
- 0
- 8
- Any value
Worked answer
4
Away from 2 the function simplifies to x+2. So the limit as x → 2 is 4. Continuity needs the value to match the limit, so k = 4. Any other choice leaves a visible dot out of place.
Problem 6
Find limx → 0 |x|x.
- It does not exist
- 0
- 1
- -1
- ∞
Worked answer
It does not exist
For x > 0, |x| = x and the quotient is 1. For x < 0, |x| = -x and the quotient is -1. The two sides head for different values, so the limit does not exist — even though both one-sided limits do.
Problem 7
Find limx → 2 x3-8x-2.
- 12
- 8
- 4
- 6
- 0
Worked answer
12
Substituting gives 00, so factor the difference of cubes. x3 - 8 = (x-2)(x2+2x+4), so the quotient is x2+2x+4. At x = 2: 4 + 4 + 4 = 12.
Problem 8
A function has limx → 1 f(x) = 5 but f(1) = 3. Which statement is true?
- The limit exists, but f is not continuous at 1
- f is continuous at 1
- The limit does not exist
- f(1) must really be 5
- Nothing can be concluded
Worked answer
The limit exists, but f is not continuous at 1
Continuity at a point needs three things: the value exists, the limit exists, and they are equal. Here the first two hold but the third fails. The graph has a single point sitting off the curve — a removable discontinuity.