Limits — JEE Main PYQs

128 previous year questions on Limits (Mathematics), newest first. Open any question for its complete step-by-step solution.

Let $x = 2$ be a root of the equation $x^{2} + p x + q = 0$ and $f (x) = {\frac{1 - c o s ( x ^{2} - 4 p x + q ^{2} + 8 q + 16 )}{( x - 2 p ) ^{4}}, 0, x \neq = 2 p x = 2 p$

Then $lim_{x \to 2 p^{+}} [f (x)]$
where [ . ] denotes greatest i …
JEE Main 2023View solution →
The set of all values of a for which $lim_{x \to a} ([x - 5] - [2 x + 2]) = 0$ , where $[α]$ denotes the greatest integer less than or equal to $α$ is equal to
JEE Main 2023View solution →
$lim_{x \to \infty} \frac{( 3 x + 1 + 3 x - 1 ) ^{6} + ( 3 x + 1 - 3 x - 1 ) ^{6}}{( x + x ^{2} - 1 ) ^{6} + ( x - x ^{2} - 1 ) ^{6}} x^{3}$
JEE Main 2023View solution →
Let $f, g$ and h be the real valued functions defined on $R$ as $f (x) = {\frac{x}{∣ x ∣}, 1, x \neq = 0 x = 0$ , …
JEE Main 2023View solution →
$lim_{n \to \infty} (\frac{1}{1 + n} + \frac{1}{2 + n} + \frac{1}{3 + n} + \dots + \frac{1}{2 n})$ is equal to :-
JEE Main 2023View solution →
The value of
$Lim_{n \to \infty} \frac{1 + 2 - 3 + 4 + 5 - 6 + \dots + ( 3 n - 2 ) + ( 3 n - 1 ) - 3 n}{2 n ^{4} + 4 n + 3 - n ^{4} + 5 n + 4}$ is :
JEE Main 2023View solution →
$lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.
JEE Main 2022View solution →
If $lim_{n \to \infty} (n^{2} - n - 1 + n α + β) = 0$ then $8 (α + β)$ is equal to :
JEE Main 2022View solution →
If $lim_{n \to \infty} \frac{( n + 1 ) ^{k - 1}}{n ^{k + 1}} [(nk + 1) + (nk + 2) + \dots + (nk + n)] = 33 \cdot lim_{n \to \infty} \frac{1}{n ^{k + 1}} \cdot [1^{k} + 2^{k} + 3^{k} + \dots + n^{k}]$ , then the integral value of k is equal to ______.
JEE Main 2022View solution →
Let $β = lim_{x \to 0} \frac{α x - ( e ^{3 x} - 1 )}{α x ( e ^{3 x} - 1 )}$ for some $α \in R$ . Then the value of $α + β$ is:
JEE Main 2022View solution →
$lim_{x \to \frac{π}{4}} \frac{8 2 - ( c o s x + s i n x ) ^{7}}{2 - 2 s i n 2 x}$ is equal to
JEE Main 2022View solution →
If $lim_{x \to 1} \frac{s i n ( 3 x ^{2} - 4 x + 1 ) - x ^{2} + 1}{2 x ^{3} - 7 x ^{2} + a x + b} = - 2$ , then the value of $(a - b)$ is equal to
JEE Main 2022View solution →
The value of $lim_{x \to 1} \frac{( x ^{2} - 1 ) s i n ^{2} ( π x )}{x ^{4} - 2 x ^{3} + 2 x - 1}$ is equal to:
JEE Main 2022View solution →
$lim_{x \to 0} \frac{c o s ( s i n x ) - c o s x}{x ^{4}}$ is equal to :
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$lim_{x \to \frac{1}{2}} \frac{s i n ( c o s ^{- 1} x ) - x}{1 - t a n ( c o s ^{- 1} x )}$ is equal to :
JEE Main 2022View solution →
Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x - 3 a ]}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :
JEE Main 2022View solution →
If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ ,
where $α, β, γ \in R$ , then which of the following is NOT correct ?
JEE Main 2022View solution →
$lim_{n \to \infty} (\frac{n ^{2}}{( n ^{2} + 1 ) ( n + 1 )} + \frac{n ^{2}}{( n ^{2} + 4 ) ( n + 2 )} + \frac{n ^{2}}{( n ^{2} + 9 ) ( n + 3 )} + \dots + \frac{n ^{2}}{( n ^{2} + n ^{2} ) ( n + n )})$ is equal to
JEE Main 2022View solution →
$lim_{x \to \frac{π}{2}} (tan^{2} x ((2 sin^{2} x + 3 sin x + 4)^{\frac{1}{2}} - (sin^{2} x + 6 sin x + 2)^{\frac{1}{2}}))$ is equal to
JEE Main 2022View solution →
Let $f (x)$ be a polynomial function such that $f (x) + f^{'} (x) + f^{''} (x) = x^{5} + 64$ . Then, the value of $lim_{x \to 1} \frac{f ( x )}{x - 1}$
JEE Main 2022View solution →
The value of $lim_{n \to \infty} \frac{1}{n} \sum_{r = 0}^{2 n - 1} \frac{n ^{2}}{n ^{2} + 4 r ^{2}}$ is:
JEE Main 2021View solution →
$lim_{x \to 0} \frac{\int _{0}^{x^{2}} ( s i n t ) d t}{x ^{3}}$ is equal to:
JEE Main 2021View solution →
$lim_{x \to 2} (\sum_{n = 1}^{9} \frac{x}{n ( n + 1 ) x ^{2} + 2 ( 2 n + 1 ) x + 4})$ is equal to:
JEE Main 2021View solution →
If the value of $lim_{x \to 0} (2 - cos x cos 2 x)^{(\frac{x + 2}{x ^{2}})}$ is equal to $e^{a}$ , then a is equal to ______.
JEE Main 2021View solution →
If $f : R \to R$ is given by $f (x) = x + 1$ , then the value of $lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + f (\frac{10}{n}) + \dots . + f (\frac{5 ( n - 1 )}{n})]$ , is :
JEE Main 2021View solution →
If $lim_{x \to 0} \frac{axe ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10, α, β, γ \in R$ , then the value of $α + β + γ$ is.......
JEE Main 2021View solution →
If $lim_{x \to \infty} (x^{2} - x + 1 - a x) = b$ , then the ordered pair $(a, b)$ is :
JEE Main 2021View solution →
$lim_{n \to \infty} (1 + \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n ^{2}})^{n}$ is equal to :
JEE Main 2021View solution →
The value of $lim_{h \to 0} 2 {\frac{3 s i n ( \frac{π}{6} + h ) - c o s ( \frac{π}{6} + h )}{3 h ( 3 c o s h - s i n h )}}$ is :
JEE Main 2021View solution →
If $U_{n} = (1 + \frac{1}{n ^{2}}) (1 + \frac{2 ^{2}}{n ^{2}})^{2} \dots (1 + \frac{n ^{2}}{n ^{2}})^{n}$ , then $lim_{n \to \infty} (U_{n})^{\frac{- 4}{n ^{2}}}$ is equal to
JEE Main 2021View solution →
If $α, β$ are the distinct roots of $x^{2} + b x + c = 0$ , then $lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ is equal to
JEE Main 2021View solution →
If $α = lim_{x \to \frac{π}{4}} \frac{t a n ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )}$ and $β = lim_{x \to 0} (cos x)^{c o t x}$ are the roots of the equation, $a x^{2} + b x - 4 = 0$ , then the ordered pair $(a, b)$ is:
JEE Main 2021View solution →
$lim_{x \to 0} \frac{s i n ^{2} ( π c o s ^{4} x )}{x ^{4}}$ is equal to:
JEE Main 2021View solution →
The value of $lim_{n \to \infty} \frac{1}{n} \sum_{j = 1}^{n} \frac{( 2 j - 1 ) + 8 n}{( 2 j - 1 ) + 4 n}$ is equal to :
JEE Main 2021View solution →
The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :
JEE Main 2021View solution →
The lowest integer which is greater than $(1 + \frac{1}{1 0 ^{100}})^{1 0^{100}}$ is .........
JEE Main 2021View solution →
The value of $lim_{x \to 0} (\frac{x}{8 1 - s i n x - 8 1 + s i n x})$ is equal to :
JEE Main 2021View solution →
Let $f : R \to R$ be a continuous function. The $lim_{x \to π /4} \frac{\frac{π}{4} \int _{2}^{s e c^{2} x} f ( x ) d x}{x ^{2} - \frac{π ^{2}}{16}}$ is equal to:
JEE Main 2021View solution →
If $lim_{x \to 0} \frac{a e ^{x} - b c o s x + c e ^{- x}}{x s i n x} = 2$ , then $a + b + c$ is equal to
JEE Main 2021View solution →
The value of the limit $Lim_{θ \to 0} \frac{t a n ( π c o s ^{2} θ )}{s i n ( 2 π s i n ^{2} θ )}$ is equal to:
JEE Main 2021View solution →
The value of
$lim_{n \to \infty} \frac{[ r ] + [ 2 r ] + \dots + [ n r ]}{n ^{2}}$ ,
Where $r$ is a non-zero real number and $[r]$ denotes the greatest integer less than or equal to $r$ , is equal to:
JEE Main 2021View solution →
If $lim_{x \to 0} \frac{s i n ^{- 1} x - t a n ^{- 1} x}{3 x ^{3}}$ is equal to $L$ , then the value of $(6 L + 1)$ is :
JEE Main 2021View solution →
$lim_{x \to 2} \frac{3 ^{x} + 3 ^{3 - x} - 12}{3 ^{- x /2} - 3 ^{1 - x}}$ is equal to ____
JEE Main 2020View solution →
$lim_{x \to 0} (\frac{3 x ^{2} + 2}{7 x ^{2} + 2})^{1/ x^{2}}$ is equal to:
JEE Main 2020View solution →
$lim_{x \to 0} \frac{\int _{0}^{x} t s i n ( 10 t ) dt}{x}$ is equal to:
JEE Main 2020View solution →
$lim_{x \to 0} \frac{x ( e ^{(1 + x^{2} + x^{4} - 1) / x} - 1 )}{1 + x ^{2} + x ^{4} - 1}$
JEE Main 2020View solution →
Let $[t]$ denote the greatest integer $\leq t$ . If for some $λ \in R - {0, 1}$ , $lim_{x \to 0} \frac{1 - x + ∣ x ∣}{λ - x + [ x ]} = L$ , then $L$ is equal to:
JEE Main 2020View solution →
If $lim_{x \to 0} {\frac{1}{x ^{8}} (1 - cos \frac{x ^{2}}{2} - cos \frac{x ^{2}}{4} + cos \frac{x ^{2}}{2} cos \frac{x ^{2}}{4})} = 2^{- k}$ , then the value of $k$ is ______.
JEE Main 2020View solution →
$lim_{x \to a} \frac{( a + 2 x ) ^{\frac{1}{3}} - ( 3 x ) ^{\frac{1}{3}}}{( 3 a + x ) ^{\frac{1}{3}} - ( 4 x ) ^{\frac{1}{3}}} (a \neq = 0)$ is equal to:
JEE Main 2020View solution →
If $α$ is the positive root of the equation, $p (x) = x^{2} - x - 2 = 0$ , then $lim_{x \to α} \frac{1 - c o s ( p ( x ))}{x + α - 4}$ is equal to:
JEE Main 2020View solution →