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.
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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.
A limit is defined as a value that a function approaches using 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:
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) $\lim\limits _{x \rightarrow-1}\left(1+x+x^2+\ldots+x^{10}\right)=1+(-1)+(-1)^2+\ldots+(-1)^{10}=1$
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:
Evaluate $\lim\limits _{x \rightarrow 2} \frac{x^3-2 x-4}{x^2-3 x+2}$
$\lim\limits _{x \rightarrow 2} \frac{x^3-2 x-4}{x^2-3 x+2}$ = $\frac{0}{0} $ (indeterminate form)
we can re - write,
$
\Rightarrow \lim\limits _{x \rightarrow 2} \frac{(x-2)\left(x^2+2 x+2\right)}{(x-2)(x-1)}
$
$(x-2)$ will cancel out in numerator and denominators
$
\Rightarrow \lim\limits _{x \rightarrow 2} \frac{\left(x^2+2 x+2\right)}{(x-1)}=10
$
This method is used when either numerator or denominator or both have fractional powers (like $\frac{1 }{ 2},\frac{1 }{ 3}$ 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,
$\begin{array}{ll}=\lim\limits _{x \rightarrow a} \frac{a+2 x-3 x}{3 a+x-4 x}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\ =\lim\limits _{x \rightarrow a} \frac{a-x}{3(a-x)}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right) & \left(\frac{0}{0} \text { form }\right) \\ =\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right) & \\ =\frac{2}{3 \sqrt{3}} & \end{array}$
Certain limits are evaluated using known results, such as:
$\lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=1 \quad$ and $\quad \lim\limits _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^x=e$
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:
$
\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 $\frac{0 }{ 0}$ form. We can rationalize the irrational powers
$\begin{aligned} & \lim\limits _{\mathbf{x}^{\rightarrow 3}} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}} \\ & =\lim\limits _{x \rightarrow 3} \frac{\sqrt{3 x}-3}{\sqrt{2 x-4}-\sqrt{2}} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{2 x-4}+\sqrt{2}} \times \frac{\sqrt{3 x}+3}{\sqrt{3 x}+3} \\ & =\lim\limits _{x \rightarrow 3} \frac{3 x-9}{2 x-4-2} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{3 x}+3} \\ & =\lim\limits _{x \rightarrow 3} \frac{3(x-3)}{2(x-3)} \times \frac{\sqrt{2 x-4}+\sqrt{2}}{\sqrt{3 x}+3} \\ & =\frac{3}{2} \cdot \frac{\sqrt{2 \times 3-4}+\sqrt{2}}{\sqrt{3 \times 3}+3} \\ & =\frac{3}{2} \times \frac{\sqrt{2}+\sqrt{2}}{6} \\ & =\frac{2 \sqrt{2}}{4}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\end{aligned}$
Hence, the answer is the option (2).
Example 5: Example 5: Let $a$ be an integer such that $\lim _{x \rightarrow 7} \frac{18-[1-x]}{[x-3 a]}$ exists, where $[\mathrm{t}]$ is greatest integer $\leqslant t$.
Then $a$ is equal to: [JEE Main 2022]
1) $-6$
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).
Evaluating the value of a function is impossible without the application of limits. The concept of limit is the cornerstone on which the development of calculus rests. Calculus has many applications in various domains like physics, biology, engineering, etc.
To evaluate the limit of an algebraic function using direct substitution, you simply substitute the value of the variable into the function. If the function is continuous at that point, the limit is equal to the function's value at that point. For example, $\lim\limits _{x \rightarrow 2}\left(x^2+3 x+5\right)=15$, by substituting $x=2$.
If direct substitution results in an indeterminate form like $\frac{0}{0}$, you can use methods such as factoring, rationalizing, or applying special limit theorems. For instance, to evaluate $\lim\limits _{x \rightarrow 1} \frac{x^2-1}{x-1}$, you factor the numerator to get $\frac{(x-1)(x+1)}{x-1}$, and then cancel the common factor to find the limit as $x+1$ when $x=1$, resulting in $2$.
For limits involving square roots or other irrational expressions, rationalizing the numerator or the denominator can be helpful. This involves multiplying by a conjugate to simplify the expression. For example,$\lim\limits _{x \rightarrow 0} \frac{\sqrt{x+4}-2}{x}$ can be evaluated by multiplying by the conjugate to simplify the expression and then substituting $x=0$ to find the limit.
Limits help in understanding the behavior of functions as they approach infinity or negative infinity. This is crucial for analyzing the long-term behavior of functions. For example, the limit $\lim\limits _{x \rightarrow \infty} \frac{1}{x}=0$ shows that as $x$ becomes infinitely large, the function $\frac{1}{x}$ approaches $0$ . This helps in understanding asymptotic behavior and horizontal asymptotes of functions.
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