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Limit of Algebraic function - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Limit of Algebraic function, Algebraic Function of type ‘infinity/infinity' is considered one of the most asked concept.
311 Questions around this concept.

Solve by difficulty

If $\lim _{x \rightarrow 0} \frac{3+\alpha \sin x+\beta \cos x+\log _e(1-x)}{3 \tan ^2 x}=\frac{1}{3}$, then $2 \alpha-\beta$ is equal to:

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Concepts Covered - 3

Limit of Algebraic function

Limit of Algebraic function

(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:

$\\\mathrm{To \;evaluate\;\lim _{x \rightarrow 2} \frac{x^{3}-2 x-4}{x^{2}-3 x+2}}\\\\\mathrm{we\;can\;re-write,}\\\\\mathrm{\Rightarrow \lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+2\right)}{(x-2)(x-1)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left [ \frac{0}{0}\;form \right ]}\\\\\mathrm{(x-2)\;will\;cancel\;out\;in\;numerator\;and\;denominators}\\\\\mathrm{\Rightarrow \lim _{x \rightarrow 2} \frac{\left(x^{2}+2 x+2\right)}{(x-1)}=10}$

(c) Rationalization Method

This method is used when either numerator or denominator or both have fractional powers (like ½, ⅓ 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:

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

Rationalizing numerator and denominator we get,

$\\=\lim _{x \rightarrow a} \frac{a+2 x-3 x}{3 a+x-4 x}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \frac{0}{0}\;form \right )\\\\=\lim _{x \rightarrow a} \frac{a-x}{3(a-x)}\left(\frac{\sqrt{3 a+x}+2 \sqrt{x}}{\sqrt{a+2 x}+\sqrt{3 x}}\right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \frac{0}{0}\;form \right )\\\\=\frac{1}{3}\left(\frac{2 \sqrt{a}+2 \sqrt{a}}{\sqrt{3 a}+\sqrt{3 a}}\right)\\\\=\frac{2}{3 \sqrt{3}}$

Limit of Algebraic Function Using Standard Result

Limit of Algebraic Function Using Standard Result

As we studied in Binomial Theorem, the binomial expansion for any index

$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3} \ldots$

where, |x| < 1

When x is infinitely small (approaching to zero), we can ignore higher powers of x and can write (1 + x)ⁿ = 1 + nx (approximately).

We will use this fact to evaluate the value of

$\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a} .$

Let x = a + h,

$\\\therefore \;\;\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=\lim_{h\rightarrow 0}\frac{(a+h)^n-a^n}{(a+h)-a}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\lim _{h \rightarrow 0} \frac{a^{n}\left\{\left(1+\frac{h}{a}\right)^{n}-1\right\}}{h}\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a^{n} \lim _{h \rightarrow 0} \frac{\left\{1+n \frac{h}{a}\right\}-1}{h}\\\\\text{[when x}\rightarrow 0, (1+x)^n\rightarrow 1+nx]\\\\\text{}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a^n\cdot\frac{n}{a}=na^{n-1}\\\\\mathbf{\\\therefore \;\;\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=na^{n-1}}\\\\\text{Also,}\;\;\;\;\lim_{x\rightarrow 0}\frac{(1+x)^n-1}{x}=n$

Illustration 1:

$\\\mathrm{If\;\;\lim_{x\rightarrow 3}\frac{x^n-3^n}{x-3}=108,\text{find the value of n such that}\;n\in\mathbb{N}}$

Solution:

$\\\mathrm{we\;have\;\lim_{x\rightarrow 3}\frac{x^n-3^n}{x-3}=108,}\\\mathrm{\Rightarrow n\cdot3^{n-1}=108=4\times3^3}\\\mathrm{\Rightarrow n=4}$

Algebraic Function of type ‘infinity/infinity'

Algebraic Function of type ‘infinity/infinity’

To find the limit of the type x ➝ ∞, write the given expression in the form of $\frac{N}{D}$ (D ≠ 0), (N is numerator, D is denominator). Then divide both N and D by highest power of x occurring in both N and D to get a meaningful form.

An important result:

$\\\text { If } m, n \text { are positive integers and } a_{0}, b_{0} \neq 0 \text { are non-zero real numbers, then }\\\\\lim _{x \rightarrow \infty} \frac{a_{0} x^{m}+a_{1} x^{m-1}+\ldots+a_{m-1} x+a_{m}}{b_{0} x^{n}+b_{1} x^{n-1}+\ldots+b_{n-1} x+b_{n}}=\begin{cases} 0& \text{ if } \;\;\;m<n \\ \frac{a_0}{b_0}& \text{ if } \;\;m=n \\ \infty & \text{ if } \;\;m>n \;\;\;when\;\;a_0b_0>0\\ -\infty & \text{ if } \;\;m>n\;\;\;when\;\;a_0b_0<0 \end{cases}$