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Limit of Indeterminate Form and Algebraic limit - Practice Questions & MCQ

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

Quick Facts

  • Limit of Indeterminate Form and Algebraic limit is considered one the most difficult concept.

  • 94 Questions around this concept.

Solve by difficulty

Let f : R\rightarrow R be a positive increasing function with \lim_{x\rightarrow \infty }\frac{f(3x)}{f(x)}=1.\; Then\; \lim_{x\rightarrow \infty }\frac{f(2x)}{f(x)}=

 Find limit  \lim _{x \rightarrow 3} \frac{\left(x^2-4 x+3\right)}{\left(x^2-5 x+6\right.}

Find Limit : \lim _{x \rightarrow 2} \frac{\left(3 x^2+2 x-5\right)}{\left(x^2-1\right)}

Find limit:  \lim _{x \rightarrow 3} \frac{\left(x^2-5 x+6\right)}{(x-3)^3}

The value of \lim _{n \rightarrow \infty} \frac{1+2-3+4+5-6+\cdots \ldots+(3 n-2)+(3 n-1)-3 n}{\sqrt{2 n^{4}+4 n+3-\sqrt{n^{4}+5 n+4}}} is:

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\mathrm{ If \lim _{x \rightarrow \infty}\left\{\frac{x^2+1}{x+1}-(\alpha x+\beta)\right\}=0, then\, \, \alpha \, \, \, and\, \, \beta \, \, are \, \, given\, \, by }

Find \lim_{\infty}(\tan x)^{\tan 2 x}:

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The value of \lim_{x\rightarrow 1}\mathrm{(1-x)\tan \left ( \frac{\pi\: x}{2} \right )} is

Is the function \mathrm{\frac{\sqrt{(1+x)}-\sqrt{(1-x)}}{x}}defined for all values of x ? Indicate the values of x for which it is defined and real. Find the limit as \mathrm{ x \rightarrow 0.}

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Evaluate  \mathrm{\operatorname{Lt}_{x \rightarrow 0} \frac{x \cdot 2^x-x}{1-\cos x}}

Concepts Covered - 1

Limit of Indeterminate Form and Algebraic limit

Limit of Indeterminate Form and Algebraic limit

If we directly substitute x = a in f(x) while evaluating \lim_{x\rightarrow a}f(x), and will get one of the seven following forms \frac{0}{0},\frac{\infty}{\infty},\infty-\infty,1^\infty,0^0,\infty^0,\infty\times0 then it is called indeterminate form.

For example,

\\(i)\;\lim_{x\rightarrow 2}\frac{x^2-4}{x-2}=\frac{0}{0}\;\text{indeterminate form.}\\(ii)\;\lim_{x\rightarrow 0}\frac{\sin x}{x}=\frac{0}{0}\;\text{indeterminate form.}\\(iii)\;\lim_{x\rightarrow \pi/2}\left (\tan x \right )^{\cos x}=\infty^0\;\text{indeterminate form.}


1. \frac{0}{0} form means the numerator and denominator are both tending to 0 (AND NOT exactly 0)

Eg, \lim_{x\rightarrow 0^+}\frac{[x]}{x}, when we input the values of x close to 0(and x>0), then denominator is tending to 0, but numerator values are not tending to 0 BUT numerator is exactly 0. So this is not \frac{0}{0} form.

2. We can convert one indeterminate form into another and vice verse.


We will divide the problems of  finding limits into five categories, which are

  1. Limit of Algebraic function

  2. Trigonometric Limit

  3. Logarithmic limit

  4. Exponential limit

  5. Miscellaneous forms


One by one we will discuss all of these.

1. Limit of Algebraic function

(a) Direct Substitution Method

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

  • If f (a) is finite then L = f (a)
  • If f (a) is undefined then limit does not exist.
  • If f (a) is indeterminate then this method fails.

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

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Limit of Indeterminate Form and Algebraic limit

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