Limits — JEE Main PYQs

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

$lim_{x \to 0} \frac{s i n ^{2} x}{2 - 1 + c o s x}$ equals
JEE Main 2019View solution →
Let $f : R \to R$ be a differentiable function satisfying $f^{'} (3) + f^{'} (2) = 0$ . Then $lim_{x \to \infty} (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{\frac{1}{x}}$ is equal to:
JEE Main 2019View solution →
For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then, $lim_{x \to 1 +} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ] )}{∣1 - x ∣∣1 - x ∣}$
JEE Main 2019View solution →
$lim_{x \to 0} \frac{x c o t ( 4 x )}{s i n ^{2} x c o t ^{2} ( 2 x )}$ is equal to:
JEE Main 2019View solution →
Let $f : R \to R$ be a continuously differentiable function such that $f (2) = 6$ and $f^{'} (2) = \frac{1}{48}$ . If $\int_{6}^{f (x)} 4 t^{3} d t = (x - 2) g (x)$ , then $lim_{x \to 2} g (x)$ is equal to:
JEE Main 2019View solution →
If $lim_{x \to 1} \frac{x ^{2} - ax + b}{x - 1} = 5$ , then $a + b$ is equal to
JEE Main 2019View solution →
Let $[x]$ denote the greatest integer less than or equal to x . Then:
$lim_{x \to 0} \frac{t a n ( π s i n ^{2} x ) + ( ∣ x ∣ - s i n ( x [ x ]) ) ^{2}}{x ^{2}}$
JEE Main 2019View solution →
$lim_{x \to π /4} \frac{c o t ^{3} x - t a n x}{c o s ( x + π /4 )}$ is:
JEE Main 2019View solution →
Let $f (x) = 5 - ∣ x - 2∣$ and $g (x) = ∣ x + 1∣$ , $x \in R$ . If $f (x)$ attains maximum value at $α$ and $g (x)$ attains minimum value at $β$ , then $lim_{x \to - α β} \frac{( x - 1 ) ( x ^{2} - 5 x + 6 )}{x ^{2} - 6 x + 8}$ is equal to :
JEE Main 2019View solution →
$lim_{x \to 0} \frac{x + 2 s i n x}{x ^{2} + 2 s i n x + 1 - s i n ^{2} x - x + 1}$ is:
JEE Main 2019View solution →
$lim_{x \to 1^{-}} \frac{π - 2 s i n ^{- 1} x}{1 - x}$ is equal to :
JEE Main 2019View solution →
$lim_{n \to \infty} (\frac{n}{n ^{2} + 1 ^{2}} + \frac{n}{n ^{2} + 2 ^{2}} + \frac{n}{n ^{2} + 3 ^{2}} + \dots + \frac{1}{5 n})$ is equal to
JEE Main 2019View solution →
If $f (x) = [x] - [\frac{x}{4}], x \in R$ , where $[x]$ denotes the greatest integer function, then:
JEE Main 2019View solution →
For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to
JEE Main 2019View solution →
$lim_{y \to 0} \frac{1 + 1 + y ^{4} - 2}{y ^{4}} =$
JEE Main 2019View solution →
If $lim_{x \to 1} \frac{x ^{4} - 1}{x - 1} = lim_{x \to k} \frac{x ^{3} - k ^{3}}{x ^{2} - k ^{2}}$ , then $k$ is
JEE Main 2019View solution →
$n \to \infty lim (\frac{( n + 1 ) ^{1/3}}{n ^{4/3}} + \frac{( n + 2 ) ^{1/3}}{n ^{4/3}} + \dots . + \frac{( 2 n ) ^{1/3}}{n ^{4/3}})$

is equal to:
JEE Main 2019View solution →
For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then
$lim_{x \to 0 +} x ([\frac{1}{x}] + [\frac{2}{x}] + \dots .. + [\frac{15}{x}])$
JEE Main 2018View solution →
$lim_{x \to \frac{π}{2}} \frac{c o t x - c o s x}{( π - 2 x ) ^{3}}$ equals :
JEE Main 2017View solution →
Let $p = lim_{x \to 0 +} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to:
JEE Main 2016View solution →
$lim_{n \to \infty} (\frac{( n + 1 ) ( n + 2 ) \dots 3 n}{n ^{2 n}})^{1/ n}$ is equal to:
JEE Main 2016View solution →
$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to :
JEE Main 2015View solution →
$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to
JEE Main 2014View solution →
$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to
JEE Main 2013View solution →
$lim_{x \to 2} (\frac{1 - c o s { 2 ( x - 2 )}}{x - 2})$
JEE Main 2011View solution →
$lim_{n \to \infty} [\frac{1}{n ^{2}} sec^{2} \frac{1}{n ^{2}} + \frac{2}{n ^{2}} sec^{2} \frac{4}{n ^{2}} + \dots . + \frac{1}{n ^{2}} sec^{2} 1]$ equals
JEE Main 2005View solution →
Let $α$ and $β$ be the distinct roots of $a x^{2} + b x + c = 0$ , then $lim_{x \to α} \frac{1 - c o s ( a x ^{2} + b x + c )}{( x - α ) ^{2}}$ is equal to
JEE Main 2005View solution →
If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of $a$ and $b$ , are
JEE Main 2004View solution →