Limit of Algebraic functions

Limits are among the most intuitive concepts in calculus, and through them, we can understand how functions relate to specific points. It allows us to understand the behavior of functions at or as the input values approach specific points or become infinitely large. Many of the algebraic functions we might form using polynomials and rational expressions cannot be carefully analyzed without understanding limits.

In this article, we shall discuss the concept of Limits of Algebraic Functions. This falls under the broader category of Calculus, which is one significant chapter within Class 11 Mathematics. It is essential not only for board exams but also for competitive exams, including Joint Entrance Examination Main and other entrance exams, namely: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of five questions have been asked on this topic in JEE Main from 2013 to 2023 including one in 2017, one in 2019, one in 2020, one in 2021, and one in 2022.

Limit of Algebraic function

A limit is defined as a value that a function approaches the output for the given input values. In other words, the limiting value of a function f(x) as x approaches a is symbolically represented by $\lim\limits _{x \rightarrow a} f(x)$ ,

that is to say when x tends toward $\mathrm{a}, \mathrm{f}(\mathrm{x})$ tends toward a constant value L .

The determination of limits of algebraic functions can be done in several ways: substitution, factoring, rationalizing, and using particular theorems about limits. Let's now discuss each one:

(a) Direct Substitution Method

To find $\lim\limits _{x \rightarrow a} f(x)$,directly substitute the value of the limit of the variable. (i.e. substitute $x=a$ ) in the expression $\mathrm{f}(\mathrm{x})$

- If $f(a)$ is finite then $L=f(a)$
- If $\mathrm{f}(\mathrm{a})$ is undefined then limit does not exist.
- If $f(a)$ is indeterminate then this method fails.

(i) $\lim\limits _{x \rightarrow 3}(x(x+1))=3(3+1)=12$
(ii) $\lim\limits _{x \rightarrow 1}\left(\frac{x^2+1}{x+100}\right)=\frac{1+1}{1+100}=\frac{2}{101}$
(iii) $\lim\limits _{x \rightarrow-1}\left(1+x+x^2+\ldots+x^{10}\right)=1+(-1)+(-1)^2+\ldots+(-1)^{10}=1$

(b) Factorization Method

In this method, we factorize numerators and denominators. The common factors are canceled out and the rest of the output is the final answer.

Illustration 1:

To evaluate $\lim\limits _{x \rightarrow 2} \frac{x^3-2 x-4}{x^2-3 x+2}$
we can re - write,

$
\Rightarrow \lim\limits _{x \rightarrow 2} \frac{(x-2)\left(x^2+2 x+2\right)}{(x-2)(x-1)} \quad\left[\frac{0}{0} \text { form }\right]
$

$(x-2)$ will cancel out in numerator and denominators

$
\Rightarrow \lim\limits _{x \rightarrow 2} \frac{\left(x^2+2 x+2\right)}{(x-1)}=10
$

(c) Rationalization Method

This method is used when either numerator or denominator or both have fractional powers (like $1 / 2,1 / 3$ etc). After rationalization, the terms are factorized which on cancellation gives the final answer.

Let’s go through an illustration to understand better

Illustration 2:

Evaluate $: \lim\limits _{x \rightarrow a} \frac{\sqrt{a+2 x}-\sqrt{3}}{\sqrt{3 a+x}-2 \sqrt{3}}$

Rationalizing numerator and denominator we get,

$\begin{array}{ll}=\lim\limits _{x \rightarrow a} \frac{a+2 x-3 x}{3 a+x-4 x}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\ =\lim\limits _{x \rightarrow a} \frac{a-x}{3(a-x)}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\ =\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right) & \\ =\frac{2}{3 \sqrt{3}} & \end{array}$

(d) Special Limit Theorems:

Certain limits are evaluated using known results, such as:

$\lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1 \quad$ and $\quad \lim\limits _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e$

Solved Examples Based On Limits of Algebraic Function:

Example 1: Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$. If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $x \rightarrow-\alpha \beta \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to :
[JEE Main 2019]
1) 0.5
2) -0.667
3) -0.5
4) 0.667

Solution:
Maxima of $f(x)_{\text {occurred at }} x=2$ i.e. $\alpha=2$
Minima of $g(x)$ occurred at $x=-1$ i.e. $\beta=-1$
By method of factorisation,

$
\lim\limits _{x \rightarrow 2} \frac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}=\frac{1}{2}
$
Hence, the answer is the option 1.

Example 2: The value of $\lim\limits _{x \rightarrow 0}\left(\frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}}\right)$ is equal to:
[JEE Main 2021]
1) 0
2) 4
3) -4
4) -1

Solution: $\square$

$
\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{x}{\sqrt[8]{1-\sin x}-\sqrt[8]{1+\sin x}} \\
& =\lim\limits _{x \rightarrow 0} x \frac{(\sqrt[8]{1-\sin x}+\sqrt[8]{1+\sin x})(\sqrt[4]{1-\sin x}+\sqrt[4]{1+\sin x})(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{((1-\sin x)-(1+\sin x))} \\
& =\lim\limits _{x \rightarrow 0} \frac{x \cdot 2 \cdot 2 \cdot 2}{-2 \sin x} \\
& =-4
\end{aligned}
$
Hence, the answer is the option 3 .
Example $3: \lim\limits _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{-\frac{x}{2}}-3^{1-x}}$ is equal to:
[JEE Main 2020]

1) 36

2) 72

3) 18

4) None of these

Solution:

Rationalizing numerator and denominator we get,

Rationalizing numerator and denominator we get,

$
\begin{aligned}
& \lim\limits _{x \rightarrow 2} \frac{3^x+3^{3-x}-12}{3^{\frac{-x}{2}}-3^{1-x}} \\
& \Rightarrow \lim\limits _{x \rightarrow 2} \frac{3^x+\frac{3^3}{3^x}-12}{\frac{1}{3^{\frac{1}{2}}}-\frac{3}{3^x}} \\
& \Rightarrow \lim\limits _{x \rightarrow 2} \frac{3^{2 x}+3^3-12 \cdot 3^x}{3^{\frac{x}{2}}-3}
\end{aligned}
$

Multiply by the conjugate of $3^{\frac{x}{2}}-3: \frac{3^x-9}{3^{\frac{x}{2}}+3}$

$
\begin{aligned}
& =\lim\limits _{x \rightarrow 2}\left(\frac{3^{2 x}+3^3-12 \cdot 3^x}{\frac{3^x-9}{3^{\frac{4}{2}+3}}}\right) \\
& =\lim\limits _{x \rightarrow 2}\left(\left(3^x-3\right)\left(3^{\frac{x}{2}}+3\right)\right)=36
\end{aligned}
$
Hence, the answer is the option (1).
Example 4: $\lim\limits ^{x \rightarrow 3} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}}$ is equal to: [JEE Main 2017]
1) $\sqrt{3}$
2) $\frac{1}{\sqrt{2}}$
3) $\frac{\sqrt{3}}{2}$

$
\frac{1}{2 \sqrt{2}}
$

Solutions:

Here we can apply the concept of Limit using the rationalization method, discussed above with an example.

The limit is of $0 / 0$ form. We can rationalize the irrational powers

$\begin{aligned} & \lim\limits _{\mathbf{F}^{\rightarrow 3}} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}} \\ & =\lim\limits _{x \rightarrow 3} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{2 x-4}+\sqrt{2}} \times \frac{\sqrt{3 x}+3}{\sqrt{3 x}+3} \\ & =\lim\limits _{x \rightarrow 3} \frac{3 x-9}{2 x-4-2} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{3 x}+3} \\ & =\lim\limits _{x \rightarrow 3} \frac{3(x-3)}{2(x-3)} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{3 x}+3} \\ & =\frac{3}{2} \cdot \frac{\sqrt{2 \times 3-4}+\sqrt{2}}{\sqrt{3 \times 3}+3} \\ & =\frac{3}{2} \times \frac{\sqrt{2}+\sqrt{2}}{6} \\ & =\frac{2 \sqrt{2}}{4}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\end{aligned}$

Hence, the answer is the option (2).

Example 5: Let be an integer such that $\lim\limits _{x \rightarrow 7} \frac{18-[1-x]}{[x-3 a]}$ exists, where is greatest integer . Then is equal to: [JEE Main 2022]

【) -6
2) -2
3) 2
4) 6

Solution:

$
\begin{aligned}
& \lim\limits _{x \rightarrow 7^{+}}=\frac{18-1-[-x]}{[2]-3 a} \\
& =\lim\limits _{x \rightarrow 7^{+}} \frac{17-(-x-1)}{x-3 a}=\lim\limits _{x \rightarrow 7^{+}} \frac{18+x}{x-3 a}=\frac{25}{7-3 a} \\
& \lim\limits _{x \rightarrow 7^{-}}=\frac{18-1-[-x]}{[x]-3 a}=\lim\limits _{x \rightarrow 7^{-}} \frac{17-(-x)}{x-1-3 a}=\frac{24}{6-3 a}
\end{aligned}
$
For limit to exist

$
\begin{aligned}
& \text { For limit to exist } \mathrm{LHL}=\mathrm{RHL} \Rightarrow \frac{25}{7-3 \mathrm{a}}=\frac{24}{6-3 \mathrm{a}} \\
& \Rightarrow 150-75 \mathrm{a}=168-72 \mathrm{a} \Rightarrow 3 \mathrm{a}=-18 \Rightarrow \mathrm{a}=-6
\end{aligned}
$
Hence the answer is the option (1).

Summary: The concept of limit is the cornerstone on which the development of calculus rests. If a problem on limits does not involve trigonometric, or inverse trigonometric then it is a problem on algebraic units, otherwise, it is a problem on algebraic limits.