MHT-CET Maths · Limits

Limits — Existence, One-Sided Limits and Limits at Infinity

A limit is where a function is heading, not where it is — so it can fail to exist when the two sides disagree, and it can be asked as x runs off to infinity.

All Limits notes MHT-CET Maths strategy

Why this matters

This is the foundation block: 17 PYQs at 35% HARD, the gentlest page of the chapter, but every later page assumes it. Two question types recur almost verbatim across sittings — a modulus or greatest-integer expression near 0 where the left and right sides must be compared, and a ratio at infinity that is settled by the leading powers alone. Learn the one-sided habit here, because the continuity pages use it on every question.

Concept 1 of 6: What a Limit Says

\lim_{x\to a} f(x) = L

is a statement about the approach, not the arrival: as

x

gets close to

a

— from either side, never equal to

a

— the values

f(x)

get close to

L

. Whether

f(a)

exists, or what it equals, is a separate question.

Definition

$\lim_{x\to a} f(x) = L$ means $f(x)$ can be made as close to $L$ as we like by taking $x$ close enough to $a$ , with $x \neq a$ .
The value $f(a)$ plays no part: it may be different from $L$ , or not exist at all.
Algebra of limits: limits of sums, differences, products and quotients are the sums, differences, products and quotients of the limits — provided each limit exists and a quotient's denominator limit is not $0$ .
Direct substitution is legal wherever it makes sense: polynomials everywhere, rational functions where the denominator is non-zero, $\sin$ , $\cos$ , $e^x$ , $\log$ and roots at any point of their domain.
The whole chapter is about the cases where substitution fails — $0/0$ , $\infty/\infty$ , $\infty-\infty$ , $1^\infty$ — and the tools that resolve each.

Algebra of limits

\lim (f \pm g) = \lim f \pm \lim g \qquad \lim (fg) = \lim f \cdot \lim g \qquad \lim \frac{f}{g} = \frac{\lim f}{\lim g}\ \ (\lim g \neq 0)

$\lim$ all limits taken as $x \to a$ , each assumed to exist

Worked example

Evaluate

\lim_{x\to 2} \dfrac{x^3 - 3x + 1}{x + 3}

Practise this concept4 quick reps

The value at the point is not the limit

f(x) = \dfrac{x^2}{x}

has no value at

x = 0

, yet

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

. Conversely a function can be defined at a point with a value that has nothing to do with its limit there. Keep the two questions separate — continuity is precisely the case where they agree.

Concept 2 of 6: One-Sided Limits and When a Limit Exists

Walk towards

a

from the left and from the right. If the two walks arrive at the same height, that height is the limit; if not, there is no limit at all. A modulus is the usual reason the two walks differ, because

|x|

means

x

on one side of

0

and

-x

on the other.

Definition

Left-hand limit $\lim_{x\to a^-} f(x)$ : the approach through values $x < a$ . Right-hand limit $\lim_{x\to a^+} f(x)$ : through values $x > a$ .
The limit exists exactly when both one-sided limits exist and are equal; their common value is the limit.
Near $0$ : $|x| = x$ for $x > 0$ and $|x| = -x$ for $x < 0$ . Near $a$ : $|x - a| = x - a$ on the right of $a$ and $-(x - a)$ on the left.
Method: whenever a stem contains $|\cdot|$ , $[\cdot]$ , a piecewise definition or a square root of something that changes sign, compute the two sides separately before saying anything.

Existence of a limit

\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L

Worked example

Does

\lim_{x\to 2}\dfrac{|x - 2|}{x - 2}

exist?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q103Moderate

Example 2 · Limits · Limits — Existence, One-Sided Limits and Limits at Infinity

The value of

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

Not every modulus makes the limit fail

\dfrac{x}{|x| + x^2}

tends to

+1

from the right and

-1

from the left — no limit. But

\dfrac{|x|}{|x| + x^2}

tends to

1

from BOTH sides, because the sign that flips upstairs also flips downstairs. Both versions have been set in the same paper series; the answer is decided by working the two sides, not by spotting a modulus.

Drill 2 more on one-sided limits and when a limit exists

Concept 3 of 6: Greatest-Integer and Sign Functions Near a Point

[x]

is a staircase: flat between integers, with a step of height

1

at each integer. Approaching an integer

n

from the left you are on the lower stair

n - 1

; from the right, on the stair

n

itself. Away from the integers there is no step, so the limit is simply the stair you are standing on.

Definition

$[x]$ = the greatest integer $\le x$ . So $[2.7] = 2$ , $[5] = 5$ , and — the one students get wrong — $[-0.3] = -1$ , because $-1$ is the largest integer not exceeding $-0.3$ .
At an integer $n$ : $\lim_{x\to n^-}[x] = n - 1$ and $\lim_{x\to n^+}[x] = n$ . The two-sided limit does not exist.
At a non-integer $a$ : $\lim_{x\to a}[x] = [a]$ , because $[x]$ is constant on an interval around $a$ .
Near $0$ from the left: $[x] = -1$ and $|x| = -x$ . Substitute these as constants and the expression usually collapses.
The sign function $\dfrac{x}{|x|}$ (also written $\operatorname{sgn} x$ ) is $+1$ for $x > 0$ and $-1$ for $x < 0$ : a two-step staircase with its single step at $0$ .

Greatest integer at an integer

\lim_{x\to n^-}[x] = n - 1, \qquad \lim_{x\to n^+}[x] = n \qquad (n \in \mathbb{Z})

Worked example

Evaluate

\lim_{x\to 2^-}\dfrac{[x] + x}{x}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q138Hard

Example 3 · Limits · Limits — Existence, One-Sided Limits and Limits at Infinity

For each

x\in\mathbb{R}

, let

[x]

represent greatest integer function, then

\lim_{x\to0^{-}}\frac{x([x]+|x|)\sin[x]}{|x|}

is equal to

[x] for a small negative x is −1, not 0

[-0.001] = -1

. Writing

0

here turns a

-\sin 1

answer into

0

, which is always one of the options. On the left of any integer

n

[x]

n - 1

— including

n = 0

Concept 4 of 6: Limits at Infinity — Compare the Leading Powers

When

x

is enormous,

x^{50}

dwarfs

x^{49}

, and a constant is invisible next to either. So only the highest power on top and the highest power below matter, and the limit is decided by comparing those two.

Definition

Ratio of polynomials: divide numerator and denominator by the highest power of $x$ present. Every term with a lower power becomes $\dfrac{c}{x^k} \to 0$ .
Degree on top smaller → limit $0$ . Degrees equal → ratio of leading coefficients. Degree on top larger → $\pm\infty$ , no finite limit.
$(2x + k)^{50}$ and $(2x)^{50}$ have the same leading term, so their ratio $\to 1$ whatever the constant $k$ — a sum of a hundred such terms over $(2x)^{50}$ therefore tends to $100$ .
A finite sum inside the limit ( $1 + 8 + 27 + \dots + n^3$ ) must be replaced by its closed form first ( $\left(\frac{n(n+1)}{2}\right)^2$ ); only then can leading powers be compared.
$e^{-x} \to 0$ , $a^{-x} \to 0$ for $a > 1$ , and $\dfrac{e^{x} - 1}{e^{x} + 1} \to 1$ : divide by the dominant exponential exactly as you would by the dominant power.

Ratio of polynomials at infinity

\lim_{x\to\infty}\frac{a_n x^n + \dots + a_0}{b_m x^m + \dots + b_0} = \begin{cases} 0, & n < m \\[2pt] \dfrac{a_n}{b_m}, & n = m \\[2pt] \pm\infty, & n > m \end{cases}

$n, m$ degrees of numerator and denominator
$a_n, b_m$ their leading coefficients

Worked example

Evaluate

\lim_{x\to\infty}\dfrac{3x^2 - 5x + 1}{2x^2 + 7}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q140Hard

Example 4 · Limits · Limits — Existence, One-Sided Limits and Limits at Infinity

\lim_{x\rightarrow \infty} \frac{(2x+ 1)^{50}+ (2x+ 2)^{50}+ (2x+ 3)^{50}+ ... + (2x+ 100)^{50}}{(2x)^{50}+ (10)^{50}}=

A sum of n terms is not 'n copies of the biggest term'

\dfrac{1 + 8 + 27 + \dots + n^3}{1 - n^4}

is NOT

\dfrac{n \cdot n^3}{-n^4} = -1

. The sum is

\left(\frac{n(n+1)}{2}\right)^2 \sim \frac{n^4}{4}

, so the limit is

-\dfrac{1}{4}

. Always write the closed form of a sum before comparing powers.

Drill 2 more on limits at infinity — compare the leading powers

Concept 5 of 6: A Finite Limit at Infinity Forces the Divergent Part to Vanish

\dfrac{x^2 + x + 1}{x + 1} - ax - b

is to settle down to a number as

x

grows, the piece that grows like

x

must be cancelled exactly by

ax

. That single observation fixes

a

; only then does

b

come into view.

Definition

Divide the rational function: $\dfrac{p(x)}{q(x)} = (\text{quotient}) + \dfrac{r(x)}{q(x)}$ , with $\deg r < \deg q$ , so the remainder term $\to 0$ .
For the limit of $\dfrac{p(x)}{q(x)} - ax - b$ to be finite, the coefficient of $x$ in (quotient $- ax - b$ ) must be zero — this determines $a$ .
The limit is then the constant left over, which determines $b$ .
Geometrically $y = ax + b$ is the oblique asymptote of the curve; the question is asking for it in disguise.

Finite limit at infinity

\frac{p(x)}{q(x)} = \alpha x + \beta + \frac{r(x)}{q(x)} \ \Rightarrow\ \lim_{x\to\infty}\left(\frac{p(x)}{q(x)} - ax - b\right) \text{ finite} \iff a = \alpha,\ \text{and the limit is } \beta - b

Worked example

\lim_{x\to\infty}\left(\dfrac{x^2 + 2x + 3}{x + 1} - ax - b\right) = 0

, find

a

and

b

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q115Hard

Example 5 · Limits · Limits — Existence, One-Sided Limits and Limits at Infinity

\lim_{x\to\infty}\left(\frac{x^2+x+1}{x+1} - ax - b\right) = 4

then

Solving for b before a

There is no equation for

b

until

a

has killed the growing term. Students who 'compare constants' first get

b

from the wrong expression. Order: divide → set the

x

-coefficient to zero → read off the constant.

Concept 6 of 6: Infinity Minus Infinity — Rationalise at Infinity

\sqrt{x^2 + 4x} - x

is a huge number minus a huge number, and the difference could be anything. Multiplying by the conjugate turns the subtraction into a division, where the leading powers can be compared as usual.

Definition

$\infty - \infty$ is indeterminate: it is never $0$ by inspection.
With square roots, multiply and divide by the conjugate: $\sqrt{A} - \sqrt{B} = \dfrac{A - B}{\sqrt{A} + \sqrt{B}}$ . The numerator usually drops to a lower degree; then divide by the highest power.
Nested roots need the trick twice — once for the outer difference, once more for the inner $\sqrt{1 + x^4} - x^2$ that appears.
For large $x$ , $\sqrt{x^2 + ax + b} \approx x + \dfrac{a}{2}$ : a shortcut worth remembering, and the reason the standard result below holds.

Root minus its leading term

\lim_{x\to\infty}\left(\sqrt{x^2 + ax + b} - x\right) = \frac{a}{2}

Worked example

Evaluate

\lim_{x\to\infty}\left(\sqrt{x^2 + 4x} - x\right)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q101Hard

Example 6 · Limits · Limits — Existence, One-Sided Limits and Limits at Infinity

\lim_{x\to\infty}x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x\sqrt2\right\}=

Subtracting infinities term by term

Writing

\sqrt{x^2 + 4x} - x \approx x - x = 0

is wrong; the answer is

2

. The lower-order terms under the root are exactly what survives after the leading parts cancel, so they cannot be dropped before rationalising.

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.

Formulas (6)

What a Limit Says
Algebra of limits
$\lim (f \pm g) = \lim f \pm \lim g \qquad \lim (fg) = \lim f \cdot \lim g \qquad \lim \frac{f}{g} = \frac{\lim f}{\lim g}\ \ (\lim g \neq 0)$
One-Sided Limits and When a Limit Exists
Existence of a limit
$\lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = \lim_{x\to a^+} f(x) = L$
Greatest-Integer and Sign Functions Near a Point
Greatest integer at an integer
$\lim_{x\to n^-}[x] = n - 1, \qquad \lim_{x\to n^+}[x] = n \qquad (n \in \mathbb{Z})$
Limits at Infinity — Compare the Leading Powers
Ratio of polynomials at infinity
$\lim_{x\to\infty}\frac{a_n x^n + \dots + a_0}{b_m x^m + \dots + b_0} = \begin{cases} 0, & n < m \\[2pt] \dfrac{a_n}{b_m}, & n = m \\[2pt] \pm\infty, & n > m \end{cases}$
A Finite Limit at Infinity Forces the Divergent Part to Vanish
Finite limit at infinity
$\frac{p(x)}{q(x)} = \alpha x + \beta + \frac{r(x)}{q(x)} \ \Rightarrow\ \lim_{x\to\infty}\left(\frac{p(x)}{q(x)} - ax - b\right) \text{ finite} \iff a = \alpha,\ \text{and the limit is } \beta - b$
Infinity Minus Infinity — Rationalise at Infinity
Root minus its leading term
$\lim_{x\to\infty}\left(\sqrt{x^2 + ax + b} - x\right) = \frac{a}{2}$

Watch out for (6)

The value at the point is not the limit
→ What a Limit Says
Not every modulus makes the limit fail
→ One-Sided Limits and When a Limit Exists
[x] for a small negative x is −1, not 0
→ Greatest-Integer and Sign Functions Near a Point
A sum of n terms is not 'n copies of the biggest term'
→ Limits at Infinity — Compare the Leading Powers
Solving for b before a
→ A Finite Limit at Infinity Forces the Divergent Part to Vanish
Subtracting infinities term by term
→ Infinity Minus Infinity — Rationalise at Infinity

Test yourself on Limits

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Limits chapter test Drill all 17 questions

Related notes

Discontinuities of [x], |x| and sgn x — the same one-sided habit, applied to continuity →