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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.

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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

If \alpha>\beta>0 \ \text{are the roots of the equation a }x^2+b x+1=0, and \lim _{x \rightarrow \frac{1}{\alpha}}\left(\frac{1-\cos \left(x^2+b x+a\right)}{2(1-a x)^2}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right), \text{then k is equal to}

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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<a<\frac{1}{2}\right)} 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}

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Algebra of Limits

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