Limits — JEE Main PYQs

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

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ .
Then …
JEE Main 2026View solution →
If $lim_{x \to 0} \frac{e ^{(a - 1) x} + 2 c o s b x + ( c - 2 ) e ^{- x}}{x c o s x - l o g _{e} ( 1 + x )} = 2$ , then $a^{2} + b^{2} + c^{2}$ is equal to :
JEE Main 2026View solution →
The value of
$lim_{x \to 0} \frac{l o g _{e} ( s e c ( e x ) \cdot s e c ( e ^{2} x ) \cdot \dots \cdot s e c ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}$
is equal to
JEE Main 2026View solution →
If $lim_{x \to 2} \frac{s i n ( x ^{3} - 5 x ^{2} + a x + b )}{( x - 1 - 1 ) l o g _{e} ( x - 1 )} = m$ , then $a + b + m$ is equal to:
JEE Main 2026View solution →
The product of all possible values of $α$ , for which $lim_{x \to 0} (\frac{1 - c o s ( α x ) c o s (( α + 1 ) x ) c o s (( α + 2 ) x )}{s i n ^{2} (( α + 1 ) x )}) = 2$ , is :
JEE Main 2026View solution →
Let $f (x) = lim_{y \to 0} \frac{( 1 - c o s ( x y )) t a n ( x y )}{y ^{3}}$ . Then the number of solutions of the equation $f (x) = sin x$ , $x \in R$ is:
JEE Main 2026View solution →
Let $lim_{x \to 2} \frac{( t a n ( x - 2 )) ( r ^{2} + ( p - 2 ) x - 2 p )}{( x - 2 ) ^{2}} = 5$ for some $r, p \in R$ . If the set of all possible values of q , such that the roots of the equation $rx^{2} - px + q = 0$ lie in ( 0,2 ), b …
JEE Main 2026View solution →
The value of $lim_{x \to 0} (\frac{x ^{2} s i n ^{2} x}{x ^{2} - s i n ^{2} x})$ is:
JEE Main 2026View solution →
If $lim_{x \to \infty} ((\frac{e}{1 - e}) (\frac{1}{e} - \frac{x}{1 + x}))^{x} = α$ , then the value of $\frac{l o g _{e} α}{1 + l o g _{e} α}$ equals :
JEE Main 2025View solution →
$lim_{x \to \infty} \frac{( 2 x ^{2} - 3 x + 5 ) ( 3 x - 1 ) ^{\frac{x}{2}}}{( 3 x ^{2} + 5 x + 4 ) ( 3 x + 2 ) ^{x}}$ is equal to:
JEE Main 2025View solution →
Let $f (x) = lim_{n \to \infty} \sum_{r = 0}^{n} (\frac{t a n ( x / 2 ^{r + 1} ) + t a n ^{3} ( x / 2 ^{r + 1} )}{1 - t a n ^{2} ( x / 2 ^{r + 1} )})$ .

Then …
JEE Main 2025View solution →
For some $a, b$ , let

$f (x) = a + \frac{s i n x}{x} a a 1 1 + \frac{s i n x}{x} 1 b b b + \frac{s i n x}{x}, x \neq = 0,$

$lim_{x \to 0} f (x) = λ + μ a + ν b$ . Then $(λ + μ + ν)^{2}$ is equal to:
JEE Main 2025View solution →
$lim_{x \to 0} cosec x (2 cos^{2} x + 3 cos x - cos^{2} x + sin x + 4)$ is
JEE Main 2025View solution →
If $lim_{x \to 0} \frac{c o s ( 2 x ) + a c o s ( 4 x ) - b}{x ^{4}}$ is finite, then $(a + b)$ is equal to :
JEE Main 2025View solution →
If $x \to 0 Lim (\frac{t a n x}{x})^{\frac{1}{x ^{2}}} = p$ , then $96 lo g_{e} p$ is equal to
JEE Main 2025View solution →
For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:
JEE Main 2025View solution →
Let f be a differentiable function on $R$ such that $f (2) = 1$ , $f^{'} (2) = 4$ . Let $lim_{x \to 0} (f (2 + x))^{3/ x} = e^{α}$ . Then the number of times the curve $y = 4 x^{3} - 4 x^{2} - 4 (α - 7) x - α$ meets x-axis is :-
JEE Main 2025View solution →
If $lim_{x \to 1^{+}} \frac{( x - 1 ) ( 6 + λ c o s ( x - 1 )) + μ s i n ( 1 - x )}{( x - 1 ) ^{3}} = - 1$ , where $λ, μ \in R$ , then $λ + μ$ is equal to
JEE Main 2025View solution →
For $t > - 1$ , let $α_{t}$ and $β_{t}$ be the roots of the equation $((t + 2)^{\frac{1}{7}} - 1) x^{2} + ((t + 2)^{\frac{1}{6}} - 1) x + ((t + 2)^{\frac{1}{21}} - 1) = 0$ .

If …
JEE Main 2025View solution →
$lim_{x \to 0^{+}} \frac{t a n ( 5 ( x ) ^{\frac{1}{3}} ) l o g _{e} ( 1 + 3 x ^{2} )}{( t a n ^{- 1} 3 x ) ^{2} ( e ^{5 (x)^{\frac{4}{3}}} - 1 )}$ is equal to
JEE Main 2025View solution →
Given below are two statements :
Statement I :
$lim_{x \to 0} (\frac{t a n ^{- 1} x + l o g _{e} \frac{1 + x}{1 - x} - 2 x}{x ^{5}}) = \frac{2}{5}$
Statement II : …
JEE Main 2025View solution →
If $lim_{x \to 1} \frac{( 5 x + 1 ) ^{1/3} - ( x + 5 ) ^{1/3}}{( 2 x + 3 ) ^{1/2} - ( x + 4 ) ^{1/2}} = \frac{m 5}{n ( 2 n ) ^{2/3}}$ , where $g cd (m, n) = 1$ , then $8 m + 12 n$ is equal to ____.
JEE Main 2024View solution →
Let $f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t$ , $x \in R$ . Then, $lim_{x \to 0} \frac{f ( x )}{x ^{3}}$ is equal to
JEE Main 2024View solution →
Let $a > 0$ be a root of the equation $2 x^{2} + x - 2 = 0$ . If $lim_{x \to \frac{1}{a}} \frac{16 ( 1 - c o s ( 2 + x - 2 x ^{2} ))}{( 1 - a x ) ^{2}} = α + β 17$ , where $α, β \in Z$ , then $α + β$ is equal to ____ .
JEE Main 2024View solution →
Let $f : (- \infty, \infty) - {0} \to R$ be a differentiable function such that $f^{'} (1) = lim_{a \to \infty} a^{2} f (\frac{1}{a})$ . Then $lim_{a \to \infty} \frac{a ( a + 1 )}{2} tan^{- 1} (\frac{1}{a}) + a^{2} - 2 lo g_{e} a$ is equal to
JEE Main 2024View solution →
$lim_{n \to \infty} \frac{( 1 ^{2} - 1 ) ( n - 1 ) + ( 2 ^{2} - 2 ) ( n - 2 ) + \dots\dots\dots + ( ( n - 1 ) ^{2} - ( n - 1 ) ) \cdot 1}{( 1 ^{3} + 2 ^{3} + \dots + n ^{2} ) - ( 1 ^{2} + 2 ^{2} + \dots + n ^{2} )}$ is equal to
JEE Main 2024View solution →
The value of
$lim_{x \to 0} 2 (\frac{1 - c o s x c o s 2 x 3 c o s 3 x \dots\dots 10 c o s 10 x}{x ^{2}})$ is ______ .
JEE Main 2024View solution →
If $α = lim_{x \to 0^{+}} (\frac{e ^{t a n x} - e ^{x}}{t a n x - x})$ and
$β = lim_{x \to 0} (1 + sin x)^{\frac{1}{2} c o t x}$ are the roots of the quadratic equation $a x^{2} + b x - e = 0$ , then 12 $lo g_{e} (a + b)$ is equal …
JEE Main 2024View solution →
Let …
JEE Main 2024View solution →
$Lim_{x \to 0} \frac{e - ( 1 + 2 x ) ^{\frac{1}{2 x}}}{x}$ is equal to :
JEE Main 2024View solution →
$lim_{x \to \frac{π}{2}} (\frac{\int _{x^{3}}^{(π /2)^{3}} ( s i n ( 2 t ^{1/3} ) + c o s ( t ^{1/3} ) ) d t}{( x - \frac{π}{2} ) ^{2}})$ is equal to :
JEE Main 2024View solution →
Let the slope of the line $45 x + 5 y + 3 = 0$ be $27 r_{1} + \frac{9 r _{2}}{2}$ for some $r_{1}, r_{2} \in R$ . Then $lim_{x \to 3} (\int_{3}^{x} \frac{8 t ^{2}}{\frac{3 r _{2} x}{2} - r _{2} x ^{2} - r _{1} x ^{3} - 3 x} d t)$ is equal to ______.
JEE Main 2024View solution →
Let $f : R \to (0, \infty)$ be strictly increasing function such that $lim_{x \to \infty} \frac{f ( 7 x )}{f ( x )} = 1$ . Then, the value of $lim_{x \to \infty} [\frac{f ( 5 x )}{f ( x )} - 1]$ is equal to
JEE Main 2024View solution →
If $lim_{x \to 0} \frac{ax ^{2} e ^{x} - b l o g _{e} ( 1 + x ) + cxe ^{- x}}{x ^{2} s i n x} = 1$ , then $16 (a^{2} + b^{2} + c^{2})$ is equal to ______ …
JEE Main 2024View solution →
The value of $lim_{n \to \infty} \sum_{k = 1}^{n} \frac{n ^{3}}{( n ^{2} + k ^{2} ) ( n ^{2} + 3 k ^{2} )}$ is :
JEE Main 2024View solution →
Let $f : [- \frac{π}{2}, \frac{π}{2}] \to R$ be a differentiable function such that $f (0) = \frac{1}{2}$ . If $lim_{x \to 0} \frac{x \int _{0}^{x} f ( t ) d t}{e ^{x^{2}} - 1} = α$ , then $8 α^{2}$ is equal to:
JEE Main 2024View solution →
If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then
$2 α - β$ is equal to :
JEE Main 2024View solution →
If $a = lim_{x \to 0} \frac{1 + 1 + x ^{4} - 2}{x ^{4}}$ and $b = lim_{x \to 0} \frac{s i n ^{2} x}{2 - 1 + c o s x}$ , then the value of $a b^{3}$ is :
JEE Main 2024View solution →
Let $f (x) = {x - 1, x is even, 2 x, x is odd, x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then …
JEE Main 2024View solution →
$lim_{x \to 0} \frac{e ^{2∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}$
JEE Main 2024View solution →
Let ${x}$ denote the fractional part of x and $f (x) = \frac{c o s ^{- 1} ( 1 - { x } ^{2} ) s i n ^{- 1} ( 1 - { x })}{{ x } - { x } ^{3}}, x \neq = 0$ . If $L$ and R respectively denotes the left hand limit and the right hand limit of $f (x)$ at $x = 0$ , then …
JEE Main 2024View solution →
$lim_{x \to \frac{π}{2}} (\frac{1}{( x - \frac{π}{2} ) ^{2}} \int_{x^{3}}^{(\frac{π}{2})^{3}} cos (\frac{1}{t ^{3}}) d t)$ is equal to
JEE Main 2024View solution →
$lim_{n \to \infty} {(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) \dots (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})}$ is equal to
JEE Main 2023View solution →
$lim_{x \to 0} ((\frac{1 - c o s ^{2} ( 3 x )}{c o s ^{3} ( 4 x )}) (\frac{s i n ^{3} ( 4 x )}{( l o g _{e} ( 2 x + 1 ) ) ^{5}}))$ is equal to
JEE Main 2023View solution →
Let $a_{1}, a_{2}, a_{3} \dots a_{n}$ be n positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then …
JEE Main 2023View solution →
If $α > β > 0$ are the roots of the equation $a x^{2} + b x + 1 = 0$ , and
$lim_{x \to \frac{1}{α}} (\frac{1 - c o s ( x ^{2} + b x + a )}{2 ( 1 - α x ) ^{2}})^{\frac{1}{2}} = \frac{1}{k} (\frac{1}{β} - \frac{1}{α})$ , then $k$ is equal to
JEE Main 2023View solution →
If $lim_{x \to 0} \frac{e ^{a x} - c o s ( b x ) - \frac{c x e ^{- c x}}{2}}{1 - c o s ( 2 x )} = 17$ , then $5 a^{2} + b^{2}$ is equal to
JEE Main 2023View solution →
Among :
(S1) : $lim_{n \to \infty} \frac{1}{n ^{2}} (2 + 4 + 6 + \dots\dots\dots + 2 n) = 1$
(S2) : …
JEE Main 2023View solution →
$lim_{t \to 0} (1^{\frac{1}{s i n ^{2} t}} + 2^{\frac{1}{s i n ^{2} t}} + \dots . + n^{\frac{1}{s i n ^{2} t}})^{s i n^{2} t}$ is equal to
JEE Main 2023View solution →
$lim_{x \to 0} \frac{48}{x ^{4}} \int_{0}^{x} \frac{t ^{3}}{t ^{6} + 1} d t$ is equal to ____.
JEE Main 2023View solution →