Algebra of Limits

What is Algebra of Limits?

Algebra of limits is a fundamental branch of calculus that provides rules and formulas to calculate limits algebraically.

It allows combining functions using limit laws for sums, differences, products, quotients, and powers.

For example, if we have two functions $f(x)$ and $g(x)$, the limit of their sum can be calculated as:
$ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) $.

Algebra of limits is particularly useful for resolving indeterminate forms like $0/0$ and for systematically solving limits of polynomials, rational functions, trigonometric functions, exponential functions, and logarithmic functions.

Importance of Limit Laws in Calculus

Limit laws are essential because they allow us to:

• Analyze function behavior near a point $x = a$.

• Break down complex limits into simpler components.

• Handle piecewise functions, continuous and discontinuous functions, and tricky indeterminate forms.

Without these laws, evaluating limits would require tedious calculations or graphical approximations.

How Algebra of Limits Helps in Evaluating Functions

By applying algebra of limits:

• Complex expressions can be simplified step by step.

• One-sided limits, indeterminate forms, and piecewise functions can be solved systematically.

• Techniques like factorization, rationalization, and the squeeze theorem become easy to apply.

Example:
$ f(x) = \frac{x^2 - 4}{x - 2} $.

Direct substitution gives $ \frac{0}{0} $, which is indeterminate. Using factorization:
$ f(x) = \frac{(x-2)(x+2)}{x-2} = x+2 $.

So:
$ \lim_{x \to 2} f(x) = 2 + 2 = 4 $.

Basic Limit Laws for Algebra of Limits

Learn the fundamental rules of limits, including sums, products, quotients, and powers, to simplify complex functions and solve limits algebraically with ease.

Limit of a Constant Function

If $c$ is a constant:
$ \lim_{x \to a} c = c $.

This also applies to one-sided limits:
$ \lim_{x \to a^-} c = \lim_{x \to a^+} c = c $.

Limit of a Sum and Difference of Functions

For functions $f(x)$ and $g(x)$:
$ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) $
$ \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) $

Example:
$ \lim_{x \to 3} [(2x+1) + (x^2-4)] = \lim_{x \to 3} (2x+1) + \lim_{x \to 3} (x^2-4) = 7 + 5 = 12 $.

Limit of a Product of Functions

$ \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) $.

Example:
$ \lim_{x \to 2} [x \cdot (x^2 + 1)] = \lim_{x \to 2} x \cdot \lim_{x \to 2} (x^2+1) = 2 \cdot 5 = 10 $.

Limit of a Quotient of Functions

If $ \lim_{x \to a} g(x) \neq 0 $:
$ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} $.

Example:
$ \lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2 $.

Limit of a Power of a Function

For $n$ a positive integer:
$ \lim_{x \to a} [f(x)]^n = (\lim_{x \to a} f(x))^n $.

Example:
$ \lim_{x \to 3} (2x)^3 = [\lim_{x \to 3} 2x]^3 = (6)^3 = 216 $.

Approach to Evaluate Limits Algebraically

Follow a clear, systematic approach to calculate limits using substitution, factorization, rationalization, and handling indeterminate forms for accurate results.

Direct Substitution Method

If $f(x)$ is continuous at $x = a$:
$ \lim_{x \to a} f(x) = f(a) $.

Example:
$ \lim_{x \to 2} (3x+1) = 3(2) + 1 = 7 $.

Using Factorization to Simplify Limits

Indeterminate forms like $0/0$ can be solved using factorization:
$ \lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4 $.

Rationalization Technique for Limits

For expressions with square roots:
$ \lim_{x \to 1} \frac{\sqrt{x+3}-2}{x-1} = \lim_{x \to 1} \frac{(\sqrt{x+3}-2)(\sqrt{x+3}+2)}{(x-1)(\sqrt{x+3}+2)} = \lim_{x \to 1} \frac{x-1}{(x-1)(\sqrt{x+3}+2)} = \lim_{x \to 1} \frac{1}{\sqrt{x+3}+2} = \frac{1}{4} $.

Handling Indeterminate Forms

Common indeterminate forms include: $0/0$, $\infty/\infty$.

Techniques include factorization, rationalization, algebraic simplification, and trigonometric identities.

Example:
$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $.

Special Techniques in Algebra of Limits

Explore advanced methods like the Squeeze Theorem, limits of piecewise functions, and trigonometric, exponential, and logarithmic limits to tackle tricky calculus problems.

Applying the Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ near $x = a$ and:
$ \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L $,
then:
$ \lim_{x \to a} f(x) = L $.

Limits of Piecewise Functions

For a piecewise function:
$ f(x) = \begin{cases} x^2, & x < 1 \ 2x + 1, & x > 1 \end{cases} $

• LHL: $ \lim_{x \to 1^-} f(x) = 1^2 = 1 $

• RHL: $ \lim_{x \to 1^+} f(x) = 2(1)+1 = 3 $

Since LHL $\neq$ RHL, the limit at $x=1$ does not exist.

Limits of Trigonometric Functions

$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $,
$ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 $.

Limits of Exponential and Logarithmic Functions

$ \lim_{x \to 0} \frac{e^x - 1}{x} = 1 $,
$ \lim_{x \to 1} \frac{\ln x}{x-1} = 1 $.

Solved Examples Based On Algebra Of Limits:

Example 1: $\lim\limits _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}$ is: [JEE Main 2019]
1) $4 \sqrt{2}$
2) $8$
3) $4$
4) $8 \sqrt{2}$

Solution:
Evaluation of Trigonometric limit -
$\lim\limits _{x \rightarrow a} \frac{\sin (x-a)}{x-a}=1$
$\lim\limits _{x \rightarrow a} \frac{\tan (x-a)}{x-a}=1$
put $\quad x=a+h$ where $h \rightarrow 0$
Then it comes

$
\begin{aligned}
& \lim\limits _{h \rightarrow 0} \frac{\sin h}{h}=\lim\limits _{h \rightarrow 0} \frac{\tan h}{h}=1 \\
& \therefore \lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \\
& \therefore \quad \lim\limits _{x \rightarrow 0} \frac{\tan x}{x}=1
\end{aligned}
$
Limit of product/quotient -

Limit of product/quotient is the product/quotient of individual limits such that

$
\begin{aligned}
& \lim\limits _{x \rightarrow a}(f(x) \cdot g(x)) \\
& =\lim\limits _{x \rightarrow a} f(x), \lim\limits _{x \rightarrow a} g(x), \text { given that } f(x) \text { and } g(x) \text { are non-zero finite values } \\
& \lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits _{x \rightarrow a} f(x)}{\lim\limits _{x \rightarrow a} g(x)}, \text { given that } f(x) \text { and } g(x) \text { are non-zero finite values }
\end{aligned}
$
Also $\lim\limits _{x \rightarrow a} k f(x)$

$
=k \lim\limits _{x \rightarrow a} f(x)
$
Using LH Rule

$
\lim\limits _{x \rightarrow \frac{\pi}{4}} \frac{3 \cot ^2 x\left(-\csc ^2 x-\sec ^2 x\right)}{-\sin \left(x+\frac{\pi}{4}\right)}=8
$
Hence, the answer is the option 2.

Example 2: $\stackrel{x \rightarrow 0}{\lim } \frac{(27+x)^{1 / 3}-3}{9-(27+x)^{2 / 3}}$ equals [JEE Main 2018]
1) $\frac{1}{3}$
2) $-\frac{1}{3}$
3)$-\frac{1}{6}$
4) $\frac{1}{6}$

Solution:

As we have learned

Limit of product/quotient -

Limit of product is the product of individual limits such that

$
\begin{aligned}
& \lim\limits _{x \rightarrow a} f(x) \cdot g(x) \\
& =\lim\limits _{x \rightarrow a} f(x) \cdot \lim\limits _{x \rightarrow a} g(x) \\
& \text { also } \lim\limits _{x \rightarrow a} k f(x) \\
& =k \lim\limits _{x \rightarrow a} f(x) \\
& \lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits _{x \rightarrow a} f(x)}{\lim\limits _{x \rightarrow a} g(x)} \\
& \text { using approximation }(1+x)^n \approx 1+n x \\
& \Rightarrow \lim\limits _{x \rightarrow 0} \frac{3\left[1+\frac{1}{3} \times \frac{x}{27}-1\right]}{9\left[1-1-\frac{x}{27} \times \frac{2}{3}\right]} \\
& =-1 / 6
\end{aligned}
$
Hence, the answer is the option 3.

Hence, the answer is the option 3.

Example 3: For each $t \in R$, let $[t]$ be the greatest integer less than or equal to $t$ then $\lim\limits _{x \rightarrow 1^{+}} \frac{(1-|x|+\sin |1-x|) \sin \left(\frac{\pi}{2}[1-x]\right)}{|1-x|[1-x]}$ [JEE Main 2019]
1) equals $1$

2) equals $0$

3) equals $-1$

4) does not exist

Solution:
Limit of productiquotient is the product/quotient of individual limits such that

$
\lim\limits _{x \rightarrow a}(f(x) \cdot g(x))
$

$=\lim\limits _{x \rightarrow a} f(x) \cdot \lim\limits _{x \rightarrow a} g(x)$, given that $f(x)$ and $g(x)$ are non-zero finite values
$\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits _{x \rightarrow a} f(x)}{\lim\limits _{x \rightarrow a} g(x)}$, given that $f(x)$ and $g(x)$ are non-zero finite values

$
\begin{aligned}
& \text { Also } \lim\limits _{x \rightarrow a} k f(x) \\
& =k \lim\limits _{x \rightarrow a} f(x)
\end{aligned}
$

Evaluation of Trigonometric limit -

$
\begin{aligned}
& \lim\limits _{x \rightarrow a} \frac{\sin (x-a)}{x-a}=1 \\
& \lim\limits _{x \rightarrow a} \frac{\tan (x-a)}{x-a}=1
\end{aligned}
$

put $x=a+h$ where $h \rightarrow 0$
Then it comes

$
\begin{aligned}
& \lim\limits _{h \rightarrow 0} \frac{\sin h}{h}=\lim\limits _{h \rightarrow 0} \frac{\tan h}{h}=1 \\
& \therefore \lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } \\
& \therefore \lim\limits _{x \rightarrow 0} \frac{\tan x}{x}=1 \\
& \lim\limits _{x \rightarrow 1^{+}} \frac{\left(1-|x|+\sin |1-x| \sin \left(\frac{\pi}{2}[1-x]\right)\right)}{|1-x|[1-x]} \\
& =\lim\limits _{x \rightarrow 1^{+}} \frac{(1-x)+\sin (x-1)}{(x-1)(-1)} \sin \left(\frac{\pi}{2}(-1)\right) \\
& =\lim\limits _{x \rightarrow 1^{+}}\left(1-\frac{\sin (x-1)}{x-1}\right)(-1)=0
\end{aligned}
$

Example 4: Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x)=\left\{\begin{array}{cl}x+a & x<0 \\ |x-1| & x \geqslant 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cl}x+1 & x<0 \\ (x-1)^2+b & x \geqslant 0\end{array}\right.$. where $a, b$ are non-negative real numbers. If (gof) $(x)$ is continuous for all $x \in R$, then $\mathrm{a}+\mathrm{b}$ is equal to $\qquad$ [JEE Main 2021]

1) $1$

2) $2$

3) $3$

4) $4$

Solution:

$
\begin{aligned}
& g[f(x)]=\left\{\begin{array}{cc}
f(x)+1 & f(x)<0 \\
(f(x)-1)^2+b & f(x) \geq 0
\end{array}\right. \\
& g[f(x)]=\left\{\begin{array}{cc}
x+a+1 & x+a<0 \& x<0 \\
|x-1|+1 & |x-1|<0 \& x \geq 0 \\
(x+a-1)^2+b & x+a \geq 0 \& x<0 \\
(|x-1|-1)^2+b & |x-1| \geq 0 \& x \geq 0
\end{array}\right. \\
& g[f(x)]=\left\{\begin{array}{cc}
x+a+1 & x \in(-\infty,-a) \& x \in(-\infty, 0) \\
|x-1|+1 & x \in[-a, \infty) \& x \in(-\infty, 0) \\
(x+a-1)^2+b & x \in \phi \\
(|x-1|-1)^2+b & x \in R \& x \in[0, \infty)
\end{array}\right.
\end{aligned}
$

$g(f(x))$ is continuous

$
\begin{array}{ll}
\text { at } x=-a & \text { at } x=0 \\
1=b+1 & (a-1)^2+b=b \\
b=0 & a=1 \\
\Rightarrow \quad a+b=1
\end{array}
$
Hence, the answer is the option 1.

Example 5: Let $f, g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as and $f(x)=\left\{\begin{array}{cl}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}, g(x)=\left\{\begin{array}{cl}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.\right.$ and $\mathrm{h}(\mathrm{x})=2[\mathrm{x}]-\mathrm{f}(\mathrm{x})$, where $[\mathrm{x}]$ is the greatest integer $\leq \mathrm{x}$. Then the value of $x \rightarrow 1 \lim g(h(x-1))$ is:
[JEE Main 2023]
1) $-1$
2) $0$
3) $\sin (1)$
4) $1$

Solution:

$
\begin{aligned}
& \lim _{\delta \rightarrow 0} g(\mathrm{~h}(-\delta)) \quad \delta>0 \\
& \lim _{\delta \rightarrow 0} g(-2+1) \\
& \Rightarrow g(-1)=1
\end{aligned}
$

RHL

$
\begin{aligned}
& \lim _{\delta \rightarrow 0} g(h(\delta)) \\
& \lim _{\delta \rightarrow 0} g(2 \times 0-1) \\
& \lim _{\delta \rightarrow 0} g(-1) \\
& \lim _{x \rightarrow 1} g(h(x-1)=1
\end{aligned}
$

Hence, the answer is the option 4.

List of topics related to Algebra of Limits

Below are the topics related to algebra of limits, from basic limit laws to different techniques, to build a strong foundation in mathematics.

Sandwich Theorem

Limits of Trigonometric Functions

Limits using expansion

Left and Right Hand Limits

Higher Order Derivatives

NCERT Resources

Access complete NCERT notes, solutions, and exemplar problems for Limits and Derivatives in one place. These resources follow the latest CBSE guidelines and help build strong conceptual clarity for board exams and entrance tests.

NCERT Notes for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Exempar Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

Practice Questions based on Algebra of Limits

Test your skills with carefully curated practice questions and solved examples on algebra of limits to improve accuracy and speed in limit evaluation.

Algebra Of Limits - Practice Question MCQ

We have shared the links below to practice questions on the related topics of algebra of limits: