NDA Maths · Limits & Continuity

One-Sided, Greatest-Integer & Modulus Limits

When the function behaves differently on the two sides of a point — a modulus, a greatest-integer step, or a piecewise rule — you must compute the left and right limits separately.

All Limits & Continuity notes NDA Maths strategy

Why this matters

These are where 'the limit doesn't exist' answers come from. The greatest-integer and modulus functions are the NDA's favourite trap: the two sides genuinely disagree, so blindly substituting gives the wrong answer.

Concept 1 of 3

Left-hand and right-hand limits

Intuition

Approach the point from below (

x\to a^-

) and from above (

x\to a^+

) separately. If the two agree, the limit is that common value; if they differ, the two-sided limit does not exist. Essential whenever the rule changes at the point.

Definition

LHL $=\lim_{x\to a^-}f(x)$ , RHL $=\lim_{x\to a^+}f(x)$ . The limit exists iff LHL $=$ RHL. For a piecewise $f$ , use the piece valid on each side; for a product/quotient of one-sided-sensitive parts, evaluate each side end-to-end.

Worked example

For

f(x)=\begin{cases}x+1,&x<0\\ x^2,&x\ge 0\end{cases}

, find

\lim_{x\to 0}f(x)

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Limits & ContinuityMODERATE

Let

f(x)=\frac{ax}{x+1}+b,\ x<1

and

\sqrt{x-1},\ 1\leq x\leq2

What is

\lim_{x\to0}f(x)

equal to?

[Q64 · Sep · 2023]

One-sided limits must AGREE for the limit to exist

If LHL

=2

and RHL

=2

the limit is

2

; but if LHL

\neq

RHL — say

1

and

-1

— the two-sided limit does not exist. Don't average them or pick one side: disagreement means DNE.

Drill 6 more on left-hand and right-hand limits

Concept 2 of 3

Limits of the greatest-integer function

Intuition

The greatest-integer function

\lfloor x\rfloor

jumps at every integer: just below

n

it equals

n-1

, at and just above

n

it equals

n

. So at an integer the one-sided limits differ by 1 — almost every

\lfloor x\rfloor

limit is a one-sided question in disguise.

Definition

At an integer $n$ : $\lim_{x\to n^-}\lfloor x\rfloor=n-1$ , $\lim_{x\to n^+}\lfloor x\rfloor=n$ — so $\lim_{x\to n}\lfloor x\rfloor$ does not exist. Between integers $\lfloor x\rfloor$ is constant. For $\lfloor g(x)\rfloor$ , track which integers $g$ crosses near the point (e.g. $\lfloor x^2\rfloor$ near $x=0$ ).

Worked example

Find

\lim_{x\to 2}\lfloor x\rfloor

(if it exists).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Limits & ContinuityEASY

f(x)=\frac{[x]}{|x|}

x\ne0

, where

[\cdot]

denotes the greatest integer function, then what is the right-hand limit of

f(x)

x=1

[Q77 · Sep · 2021]

⌊x⌋ jumps at integers — the limit there does NOT exist

Approaching an integer

n

\lim_{x\to n^-}\lfloor x\rfloor=n-1

while

\lim_{x\to n^+}\lfloor x\rfloor=n

. The off-by-one is the trap — just below

3

\lfloor x\rfloor=2

, not

3

. Between integers

\lfloor x\rfloor

is constant, so the limit exists only at non-integers.

Drill 6 more on limits of the greatest-integer function

Concept 3 of 3

Limits involving the modulus

Intuition

A modulus splits at its zero:

|x|=x

for

x\ge 0

and

-x

for

x<0

. Near that split the sign flips, so

x/|x|

and similar ratios have different one-sided values. Watch for hidden moduli from square roots like

\sqrt{1-\cos\theta}=\sqrt2\,|\sin\tfrac\theta2|

Definition

Replace $|g(x)|$ by $+g$ on the side where $g>0$ and $-g$ where $g<0$ , then take each one-sided limit. $\dfrac{x}{|x|}$ is $+1$ for $x>0$ and $-1$ for $x<0$ . A surd hides a modulus: $\sqrt{A^2}=|A|$ , which is sign-sensitive.

Worked example

Find

\lim_{x\to 5}\dfrac{5-x}{|x-5|}

(if it exists).

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Limits & ContinuityEASY

What is

\lim_{x\to5}\dfrac{5-x}{|x-5|}

equal to?

[Q99 · Apr · 2023]

x / |x| is +1 on the right, −1 on the left

Because

|x|=x

for

x>0

and

-x

for

x<0

\dfrac{x}{|x|}=+1

x\to0^+

and

-1

x\to0^-

. The two-sided

\lim_{x\to0}\dfrac{x}{|x|}

therefore does not exist — it is never

0

1

A square root hides a modulus

\sqrt{A^2}=|A|

, not

A

. So

\sqrt{1-\cos\theta}=\sqrt2\,\bigl|\sin\tfrac\theta2\bigr|

, which is sign-sensitive — the right and left limits of

\dfrac{\sqrt{1-\cos\theta}}{\theta}

come out

\pm\tfrac{1}{\sqrt2}

Drill 1 more on limits involving the modulus

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Watch out for (4)

One-sided limits must AGREE for the limit to exist
→ Left-hand and right-hand limits
⌊x⌋ jumps at integers — the limit there does NOT exist
→ Limits of the greatest-integer function
x / |x| is +1 on the right, −1 on the left
→ Limits involving the modulus
A square root hides a modulus
→ Limits involving the modulus

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