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)

LHL $=\lim_{x\to0^-}(x+1)=1$ ; RHL $=\lim_{x\to0^+}x^2=0$ .
LHL $\neq$ RHL.

Answer:The limit does not exist.

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

\lim_{x\to 0}\dfrac{x^2+x+|x|}{x}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Two-sided limit exists iff?
2.
Notation for the right-hand limit?
3.
If LHL $=2$ , RHL $=2$ , limit?
4.
If LHL $=1$ , RHL $=-1$ , limit?

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]

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).

LHL $=\lim_{x\to2^-}\lfloor x\rfloor=1$ ; RHL $=\lim_{x\to2^+}\lfloor x\rfloor=2$ .
They differ.

Answer:Does not exist (LHL

=1

, RHL

=2

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

\lim_{x\to 0^-}\dfrac{\lfloor x\rfloor}{|x|}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$\lim_{x\to3^-}\lfloor x\rfloor$ ?
2.
$\lim_{x\to3^+}\lfloor x\rfloor$ ?
3.
Does $\lim_{x\to n}\lfloor x\rfloor$ exist at an integer $n$ ?
4.
$\lfloor x\rfloor$ on $(2,3)$ ?

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]

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).

For $x>5$ : $|x-5|=x-5$ , ratio $=\dfrac{-(x-5)}{x-5}=-1$ . For $x<5$ : $|x-5|=5-x$ , ratio $=+1$ .
RHL $=-1$ , LHL $=+1$ .

Answer:Does not exist (the two sides give

\pm 1

Practice this conceptself-check · 4 quick reps

Try it yourself

Evaluate

\lim_{\theta\to 0^+}\dfrac{\sqrt{1-\cos\theta}}{\theta}

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$x/|x|$ for $x>0$ ?
2.
$x/|x|$ for $x<0$ ?
3.
$\sqrt{A^2}=?$
4.
Does $\lim_{x\to0}\dfrac{x}{|x|}$ exist?

From the bank · past-year question

Example 3Limits & ContinuityEASY

What is

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

equal to?

[Q99 · Apr · 2023]

Drill 1 more on limits involving the modulus

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Limits & ContinuityEASY

Which one of the following is correct regarding

\displaystyle\lim_{x\to3}\dfrac{|x-3|}{x-3}

[Q74 · Sep · 2024]

Example 2Limits & ContinuityMODERATE

Consider the following for the items that follow: Let

f(x)=\sin[\pi^{2}]x+\cos[-\pi^{2}]x

where

[\cdot]

is a greatest integer function.

What is

f\!\left(\dfrac{\pi}{2}\right)

equal to?

[Q79 · Apr · 2023]

Example 3Limits & ContinuityMODERATE

f(x) = \frac{x^2+x+|x|}{x}

, then what is

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

equal to?

[Q86 · Sep · 2022]

Example 4Limits & ContinuityMODERATE

For the following two (02) items: Let

f(x)=\begin{cases}\dfrac{1-\cos2x}{x^2}, & x<0\\ 9, & x=0\\ \dfrac{\sqrt{x}}{\sqrt{(16+\sqrt{x})}-4}, & x>0\end{cases}

What is

\displaystyle\lim_{x\to 0^+}f(x)

equal to?

[Q72 · Sep · 2025]

Example 5Limits & ContinuityHARD

Let

f(x)=|x|+1

g(x)=[x]-1

h(x)=f(x)\cdot g(x)

. What is

\displaystyle\lim_{x\to0^-}h(x)+\lim_{x\to0^+}h(x)

equal to?

[Q88 · Apr · 2024]

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