The algebra of limits is one of the core topics in learning how to manipulate and combine functions at their limits. An understanding of calculus begins with the mastery of the algebra of limits. This is elementary knowledge to consider changes, optimize processes, and forecast trends in engineering, physics, and economics. It is from this algebra of limits that we can explore the behavior of a function in the vicinity of some particular point, even though, at one specific point, the function remains undefined. This simple idea enables us to solve the most complicated problems related to the real world and equations in mathematics, which include finding derivatives and integrals.
JEE Main 2025: Sample Papers | Mock Tests | PYQs | Study Plan 100 Days
JEE Main 2025: Maths Formulas | Study Materials
JEE Main 2025: Syllabus | Preparation Guide | High Scoring Topics
In this article, we will cover the concept of the Algebra of Limits. This topic falls under the broader category of Calculus, which is a crucial chapter in Class 11 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of ten questions have been asked on this topic in JEE Main from 2013 to 2023, two questions in 2018 and three in 2019, one in 2021, and four in 2024.
Limit of a function
If $x \rightarrow a, f(x) \rightarrow l$, then $l$ is called limit of the function $f(x)$ which is symbolically written as $\lim _{x \rightarrow a} f(x)=l$.
Algebra of limits operates under several rules or properties, making it possible to determine the limits of functions from their algebraic combinations. These rules make finding limits easier and are some of the most important tools within calculus. The basic algebraic operations include addition, subtraction, multiplication, and division of limits.
Let $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ be defined for all $x \neq$ $a$ over some open interval containing $a$. Assume that L and M are real numbers such that $x \rightarrow a$ then, $\lim\limits _{x \rightarrow a} f(x)=L$ and $\lim\limits _{x \rightarrow a} g(x)=M$. Then, each of the following statements hold:
Sum law for limits : $\lim\limits _{x \rightarrow a}(f(x)+g(x))=\lim\limits _{x \rightarrow a} f(x)+\lim\limits _{x \rightarrow a} g(x)=L+M$
Difference law for limits : $\lim\limits _{x \rightarrow a}(f(x)-g(x))=\lim\limits _{x \rightarrow a} f(x)-\lim\limits _{x \rightarrow a} g(x)=L-M$
Constant multiple law for limits : $\lim\limits _{x \rightarrow a} c f(x)=c \cdot \lim\limits _{x \rightarrow a} f(x)=c L$
Product law for limits : $\lim\limits _{x \rightarrow a}(f(x) \cdot g(x))=\lim\limits _{x \rightarrow a} f(x) \cdot \lim\limits _{x \rightarrow a} g(x)=L \cdot M$
Quotient law for limits : $\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim\limits _{x \rightarrow a} f(x)}{\lim\limits _{x \rightarrow a} g(x)}=\frac{L}{M}$ for $M \neq 0$
Power law for limits : $\lim\limits _{x \rightarrow a}(f(x))^n=\left(\lim\limits _{x \rightarrow a} f(x)\right)^n=L^n$ for every positive integer $n$
Composition law of limit: $\lim\limits _{x \rightarrow a}(f \circ g)(x)=f\left(\lim\limits _{x \rightarrow a} g(x)\right)=f(M)$, only if $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{g}(\mathrm{x})=\mathrm{M}$.
If $\lim\limits _{x \rightarrow a} f(x)=+\infty$ or $-\infty$, then $\lim\limits _{x \rightarrow a} \frac{1}{f(x)}=0$
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.
The basic ideas behind the Algebra of Limits are relatively straightforward in calculus, as it offers the necessary tools that one could use to evaluate the limit of more complicated functions. Understandably, such knowledge makes the study of continuous functions and their behaviors quite an easy venture since it empowers one to gain the background required for other topics, such as derivatives and integrals. This all translates into being a fundamental concept for success not only academically in mathematics but also practically in numerous scientific and engineering applications.
The concept of limits is the cornerstone on which the development of calculus rests. Limits help in understanding the behavior of functions as they approach specific points, which is essential for studying continuity, derivatives, and integrals.
No, the limit does not exist for zero because for saying that limit exists; the function has to approach the same value regardless of which direction $x$ comes from.
A function $f(x)$ has a limit $L$ at $x=a $ if and only if it has both left and right limits at that point, and these one-sided limits are equal to Formally, $\lim\limits _{x \rightarrow a} f(x)=L$ if and only if $\lim\limits _{x \rightarrow a^{-}} f(x)=\lim\limits _{x \rightarrow a^{+}} f(x)=L$
The basic algebraic operations include addition, subtraction, multiplication, and division of limits.
$\lim\limits _{x \rightarrow a}(f(x)+g(x))=\lim\limits _{x \rightarrow a} f(x)+\lim\limits _{x \rightarrow a} g(x)=L+M$
15 Oct'24 10:01 AM
15 Oct'24 09:58 AM
15 Oct'24 09:53 AM
15 Oct'24 09:50 AM
15 Oct'24 09:47 AM
15 Oct'24 09:42 AM
15 Oct'24 09:37 AM
15 Oct'24 09:33 AM
15 Oct'24 09:19 AM
15 Oct'24 09:14 AM