JEE Main Syllabus 2025 PDF - Subject Wise Topics

# Algebra of Limits - Practice Questions & MCQ

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

## Quick Facts

• Algebra of Limits is considered one of the most asked concept.

• 54 Questions around this concept.

## Solve by difficulty

The value of the $\lim _{t \rightarrow 1} \frac{t-t^2}{\sqrt{1+\log _2 t}-1}$ is equal to:

Using the power law for limits evaluate $\lim_{x\rightarrow -2}\left ( \frac{x^{3}-2}{x^{3}+14x^{2}+2} \right )^{2}$.

The set of all values of a for which $\lim _{x \rightarrow a}([x-5]-[2 x+2])=0$ where $[\propto]$ denotes the greatest integer less than or equal to $\alpha$ is equal to

Let $f, g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as
$f(x)=\left\{\begin{array}{cl}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}, g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.\right.$
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 $\lim _{x \rightarrow 1} g(h(x-1))$ is

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Let $\mathrm{a_{n+1}=\sqrt{\left(2+a_n\right)}, n=1,2,3, \ldots \ldots\: and\: \: a_1=3}$, then the value of $\mathrm{\lim _{n \rightarrow a} a_n}$  is............

If $\mathrm{f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2}$, then the value of $\mathrm{\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}}$ is equal to .......

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If $\mathrm{f(x)=\cot ^{-1}\left(\frac{2 x}{1-x^2}\right) and \: g(x)=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right), then\: \: \lim _{x \rightarrow a} \frac{f(x)-f(a)}{g(x)-g(a)}\left(0 is ......

Let $f: R \rightarrow[0, \infty)$ be such that $\lim _{x \rightarrow 5} f(x)$ exists and $\lim _{x \rightarrow 5} \frac{f(x)^2-9}{\sqrt{|x-5|}}=0$ Then $\lim _{x \rightarrow 5} f(x)$ equals to :

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## Concepts Covered - 1

Algebra of Limits

Algebra of Limits

Let f(x) and g(x) be defined for all x ≠ a over some open interval containing a. Assume that L and M are real numbers such that $\lim _{x \rightarrow a} f(x)=L$   and $\lim _{x \rightarrow a} g(x)=M$. Let c be a constant. Then, each of the following statements hold:

$\\\mathrm{ { \mathbf{Sum \;law \;for \;limits}: }} \lim _{x \rightarrow a}(f(x)+g(x))=\lim _{x \rightarrow a} f(x)+\lim _{x \rightarrow a} g(x)=L+M\\\\\mathrm { \mathbf{Difference\;law\; for\; limits: }} \lim _{x \rightarrow a}(f(x)-g(x))=\lim _{x \rightarrow a} f(x)-\lim _{x \rightarrow a} g(x)=L-M\\\\\mathrm { \mathbf{Constant\; multiple\; law \;for\; limits:} } \lim _{x \rightarrow a} c f(x)=c \cdot \lim _{x \rightarrow a} f(x)=c L\\\\\mathrm { \mathbf{Product \;law\; for \;limits:} } \lim _{x \rightarrow a}(f(x) \cdot g(x))=\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x)=L \cdot M$

$\\\mathrm { \mathbf{Quotient\; law \;for \;limits: }} \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)}=\frac{L}{M} \text { for } M \neq 0\\\\\\\mathrm { \mathbf{Power \;law \;for \;limits:} } \lim _{x \rightarrow a}(f(x))^{n}=\left(\lim _{x \rightarrow a} f(x)\right)^{n}=L^{n} \text { for every positive integer } n\\\\\mathrm{\mathbf{Composition\;law\;of\;limit:}}\lim_{x\rightarrow a}(fog)(x)=f\left ( \lim_{x\rightarrow a}g(x) \right )=f(M), \text{only if f(x) is }\\\text{continuous at g(x)=M.}\\\\\mathrm{\mathbf{If}\;{lim_{x\rightarrow a}f(x)=+\infty}\;\mathbf{or}-\infty,\;\mathbf{then}\;\lim_{x\rightarrow a}\frac{1}{f(x)}=0}$

## Study it with Videos

Algebra of Limits

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## Books

### Reference Books

#### Algebra of Limits

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.5

Line : 17