Complete the table using a calculator and use the result to estimate the limit.
limx→2x2−4x−2
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| f(x) |
Concept understanding — Concept of Limit
What a Limit Really Asks
A limit answers one question: as the input x crowds in on some point x0 from both sides, what value does f(x) crowd in on? Crucially, this is a statement about the neighbourhood of x0, not about the point itself. Whether f is even defined at x0 — or what value it takes there if it is — is irrelevant to whether the limit exists.
Definition 9.1 (informal). Let I be an open interval containing x0 and let f:I→R. We say limx→x0f(x)=L if, whenever x is taken sufficiently close to x0 (from either side, with x=x0), f(x) gets correspondingly close to L.
A good mental model: imagine feeding the function a sequence of x-values that homes in on x0 — never landing exactly on x0. If every such approach forces f(x) toward one and the same number L, no matter how you approach, then L is the limit.
A worked illustration. Take f(x)=x2+1 near x=3. Plugging in values like 2.9,2.99,2.999 and 3.1,3.01,3.001 shows f(x) creeping toward 10 from both sides — and indeed f(3)=10 too. Here the limit and the function value agree, which is the friendly case. But that agreement is a bonus, not a requirement of the definition.
One-sided limits
Because "approaching from either side" is doing real work in the definition, it helps to split it into two directions.
- Left-hand limit (Definition 9.2): limx→x0−f(x)=l1 — the value f(x) settles toward as x increases up to x0 while staying less than x0.
- Right-hand limit (Definition 9.3): limx→x0+f(x)=l2 — the value f(x) settles toward as x decreases down to x0 while staying greater than x0.
The reconciliation theorem: the two-sided limit limx→x0f(x) exists and equals L if and only if both one-sided limits exist and agree:
limx→x0−f(x)=L=limx→x0+f(x)
If even one of the one-sided limits fails to exist, or the two one-sided limits disagree, the two-sided limit simply does not exist — there is no "in-between" value to fall back on.
Reading limits from tables and graphs
Before any algebraic machinery, limits can be estimated two ways:
- Table of values — plug in x-values that approach x0 from below and above, and watch what f(x) trends toward.
- Graph — trace the curve with your eye from both sides of x0 and see where the two traces converge (or fail to).
When limits genuinely fail to exist
The classic culprit is a function that behaves differently on the two sides of the point.
Example. Consider g(x)=∣x∣x (undefined at x=0). For x>0, ∣x∣x=1; for x<0, ∣x∣x=−1. So
limx→0−g(x)=−1,limx→0+g(x)=+1.
These disagree, so limx→0g(x) does not exist — even though g is perfectly well-behaved (constant, even) on either side individually. This is the signature pattern of a "jump" in behaviour: two honest, finite one-sided limits that simply don't match.
Similarly, a function like the greatest-integer (floor) function has a limit that fails to exist at every integer, for exactly this reason: approaching an integer n from below gives n−1, from above gives n.
Do not confuse "the limit does not exist" with "the function blows up." A function can be perfectly finite on both sides of a point and still have no limit there — it just needs to approach different finite values from the two sides.
The independence of f(x0) from the limit
This is the single most important conceptual point in the whole idea of a limit, and it cuts two ways:
- The limit can exist even when f(x0) does not. For instance, f(x)=x+4x2−16 is undefined at x=−4 (division by zero), yet for every x=−4 it simplifies to f(x)=x−4, so as x→−4 from either side, f(x)→−8. The graph has a "hole" at x=−4, but the limit exists perfectly well there.
- The limit can exist and disagree with f(x0), or f(x0) can exist while the limit doesn't. The value of the function at the point plays no role whatsoever in computing limx→x0f(x) — that computation only ever looks at points near x0, deliberately excluding x0 itself.
This independence is precisely what makes limits the right tool to define continuity later (Concept 5): continuity is the special, well-behaved case where the limit and the function value do happen to coincide. Until then, always evaluate a limit by looking at the neighbourhood, never by simply "plugging in."
Finally, note that "the limit is a real number L" is itself part of the claim — a limit is only said to exist when that number is unique and finite. If the one-sided approach values run off unboundedly (as happens with 1/x2 near 0) or never settle down at all, we again say the limit does not exist, even though we may still describe that runaway behaviour using the symbol ∞ (a topic for later).
x2−4 is a difference of squares: x2−4=(x−2)(x+2). Cancel x−2, then substitute x=2.
x→2limx2−4x−2=41
Step 1. Try direct substitution. At x=2, x2−4x−2=00, indeterminate.
Step 2. Factor. x2−4=(x−2)(x+2), so for x=2,
x2−4x−2=(x−2)(x+2)x−2=x+21.
Step 3. Table (using f(x)=x+21 for x=2).
| x | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| f(x) | 0.2564 | 0.2506 | 0.2501 | 0.2499 | 0.2494 | 0.2439 |
Both sides trend to 0.25.
Step 4. Confirm algebraically. x→2limx+21=41.
x→2limx2−4x−2=41
- Not recognizing x2−4 as a difference of squares
- Cancelling incorrectly (e.g. cancelling x instead of the common factor x−2)
- Plugging x=2 into the original unfactored expression and stopping at 0/0
- CBSE 2025Set ANNUAL1 markMCQQ.x→3lim⌊x⌋=(a) Value does not exist(b) 2(c) 0(d) 3
›Reveal solutionSolution
The greatest integer function jumps at every integer, so its two-sided limit at x=3 fails to exist.
As x→3− (values just below 3, e.g. 2.9), ⌊x⌋=2, so the left-hand limit is 2.
As x→3+ (values just above 3, e.g. 3.1), ⌊x⌋=3, so the right-hand limit is 3.
Since the left and right-hand limits are different, x→3lim⌊x⌋ does not exist.
✓Final answerThe correct option is (a) Value does not exist.
- CBSE 2019Set ANNUAL1 markMCQQ.For the function f(x)={x+2,x−2,x>0x<0(a) x→2−limf(x)=−1(b) x→0limf(x) does not exist(c) x→0−limf(x)=−1(d) x→0+limf(x)=1
›Reveal solutionSolution
As x→0+, f(x)=x+2→2; as x→0−, f(x)=x−2→−2. Since these one-sided limits differ, the overall limit at x=0 does not exist, matching option (b); the other three stated values are each incorrect when checked directly.
Check (a): limx→2−f(x) — near x=2 we're on the x>0 branch, so f(x)=x+2→2+2=4, not −1. False.
Check (b): right-hand limit at 0 is limx→0+(x+2)=2; left-hand limit at 0 is limx→0−(x−2)=−2. Since 2=−2, the two-sided limit limx→0f(x) does not exist. True.
Check (c): limx→0−f(x)=−2, not −1. False.
Check (d): limx→0+f(x)=2, not 1. False.
✓Final answerThe correct option is (b) x→0limf(x) does not exist.
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