Evaluation of Limits & L'Hopital's Rule

Interactive Feed

JEE Main 2016

LEVELBoard

Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

(A)

- π

(B)

π

(C)

π /2

(D)

JEE Main 2013

LEVELBoard

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 2} \frac{1 - c o s { 2 ( x - 2 )}}{x - 2}$

(A)

equals

2

(B)

equals

- 2

(C)

equals

1/ 2

Let $f : R \to R$ be a positive increasing function with $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$ . Then $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} =$

The function $f : R / {0} \to R$ given by $f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}$ can be made continuous at $x = 0$ by defining $f (0)$ as

Let $f : R \to [0, \infty)$ be such that $lim_{x \to 5} f (x)$ exists and $lim_{x \to 5} \frac{( f ( x ) ) ^{2} - 9}{∣ x - 5∣} = 0$ . Then $lim_{x \to 5} f (x)$ equals

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

(A)

\frac{a ^{2}}{2} (α - β)^{2}

(B)

(C)

- \frac{a ^{2}}{2} (α - β)^{2}

(D)

\frac{1}{2} (α - β)^{2}

JEE Main 2004

LEVELBoard

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

If $lim_{x \to 0} \frac{l o g ( 3 + x ) - l o g ( 3 - x )}{x} = k$ , the value of $k$ is

The value of $lim_{x \to 0} \frac{\int _{0}^{x^{2}} s e c ^{2} t d t}{x s i n x}$ is

Let $f (a) = g (a) = k$ and their $n$ th derivatives $f^{n} (a), g^{n} (a)$ exist and are not equal for some $n$ . Further if $lim_{x \to a} \frac{f ( a ) g ( x ) - f ( a ) - g ( a ) f ( x ) + f ( a )}{g ( x ) - f ( x )} = 4$ then the value of $k$ is

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

(A)

\infty

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

$lim_{x \to 0} \frac{l o g x ^{n} - [ x ]}{[ x ]}, n \in N$ , ( $[x]$ denotes greatest integer less than or equal to $x$ )

If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

If $lim_{x \to \infty} (\frac{x ^{2} + x + 1}{x + 1} - a x - b) = 4$ , then

(A)

a = 1, b = 4

(B)

a = 1, b = - 4

(C)

a = 2, b = - 3

(D)

a = 2, b = 3

JEE Advanced 2012

LEVELJEE Main

Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

(A)

- 5/2

and 1

(B)

- 1/2

and

- 1

(C)

- 7/2

and 2

(D)

- 9/2

and 3

JEE Advanced 2011

LEVELJEE Advanced

If $lim_{x \to 0} [1 + x ln (1 + b^{2})]^{1/ x} = 2 b sin^{2} θ, b > 0$ and $θ \in (- π, π]$ , then the value of $θ$ is

(A)

\pm \frac{π}{4}

(B)

\pm \frac{π}{3}

(C)

\pm \frac{π}{6}

(D)

\pm \frac{π}{2}

JEE Advanced 2010

LEVELJEE Main

The value of $lim_{x \to 0} \frac{1}{x ^{3}} \int_{0}^{x} \frac{t l n ( 1 + t )}{t ^{4} + 4} d t$ is

Let $L = lim_{x \to 0} \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}, a > 0$ . If $L$ is finite, then

* Multiple Correct Options

(A)

a = 2

(B)

a = 1

(C)

L = 1/64

(D)

L = 1/32

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let p=limx→0+​(1+tan2x​)2x1​ then logp is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2sin(πcos2x)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​xtan4x(1−cos2x)(3+cosx)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→2​x−21−cos{2(x−2)}​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be a positive increasing function with limx→∞​f(x)f(3x)​=1. Then limx→∞​f(x)f(2x)​=

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The function f:R/{0}→R given by f(x)=x1​−e2x−12​ can be made continuous at x=0 by defining f(0) as

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→[0,∞) be such that limx→5​f(x) exists and limx→5​∣x−5∣​(f(x))2−9​=0. Then limx→5​f(x) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β be the distinct roots of ax2+bx+c=0, then limx→α​(x−α)21−cos(ax2+bx+c)​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→∞​(1+xa​+x2b​)2x=e2, then the values of a and b, are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​xlog(3+x)−log(3−x)​=k, the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limx→0​xsinx∫0x2​sec2tdt​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→2π​​1+tan(x/2)1−tan(x/2)​[π−2x]3[1−sinx]​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The positive integer just greater than (1+0.0001)10000 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​2​x1−cos2x​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→∞​(x2+x+3x2+5x+3​)x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=4 and f′(x)=4. Then limx→2​x−2xf(2)−2f(x)​ is given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​[x]logxn−[x]​,n∈N, ([x] denotes greatest integer less than or equal to x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(1)=1,f′(1)=2, then limx→1​x​−1f(x)​−1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→∞​(x+1x2+x+1​−ax−b)=4, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α(a) and β(a) be the roots of the equation (31+a​−1)x2+(1+a​−1)x+(61+a​−1)=0 where a>−1. Then lima→0+​α(a) and lima→0+​β(a) are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​[1+xln(1+b2)]1/x=2bsin2θ,b>0 and θ∈(−π,π], then the value of θ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limx→0​x31​∫0x​t4+4tln(1+t)​dt is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L=limx→0​x4a−a2−x2​−4x2​​,a>0. If L is finite, then

Let $p = lim_{x \to 0^{+}} (1 + tan^{2} x)^{\frac{1}{2 x}}$ then $lo g p$ is equal to

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 0} \frac{s i n ( π c o s ^{2} x )}{x ^{2}}$ is equal to

$lim_{x \to 0} \frac{( 1 - c o s 2 x ) ( 3 + c o s x )}{x t a n 4 x}$ is equal to

$lim_{x \to 2} \frac{1 - c o s { 2 ( x - 2 )}}{x - 2}$

Let $f : R \to R$ be a positive increasing function with $lim_{x \to \infty} \frac{f ( 3 x )}{f ( x )} = 1$ . Then $lim_{x \to \infty} \frac{f ( 2 x )}{f ( x )} =$

The function $f : R / {0} \to R$ given by $f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}$ can be made continuous at $x = 0$ by defining $f (0)$ as

Let $f : R \to [0, \infty)$ be such that $lim_{x \to 5} f (x)$ exists and $lim_{x \to 5} \frac{( f ( x ) ) ^{2} - 9}{∣ x - 5∣} = 0$ . Then $lim_{x \to 5} f (x)$ equals

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

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

If $lim_{x \to 0} \frac{l o g ( 3 + x ) - l o g ( 3 - x )}{x} = k$ , the value of $k$ is

The value of $lim_{x \to 0} \frac{\int _{0}^{x^{2}} s e c ^{2} t d t}{x s i n x}$ is

$lim_{x \to \frac{π}{2}} \frac{1 - t a n ( x /2 )}{1 + t a n ( x /2 )} \frac{[ 1 - s i n x ]}{[ π - 2 x ] ^{3}}$ is

The positive integer just greater than $(1 + 0.0001)^{10000}$ is

$lim_{x \to 0} \frac{1 - c o s 2 x}{2 x}$

$lim_{x \to \infty} (\frac{x ^{2} + 5 x + 3}{x ^{2} + x + 3})^{x}$

Let $f (x) = 4$ and $f^{'} (x) = 4$ . Then $lim_{x \to 2} \frac{x f ( 2 ) - 2 f ( x )}{x - 2}$ is given by

$lim_{x \to 0} \frac{l o g x ^{n} - [ x ]}{[ x ]}, n \in N$ , ( $[x]$ denotes greatest integer less than or equal to $x$ )

If $f (1) = 1, f^{'} (1) = 2$ , then $lim_{x \to 1} \frac{f ( x ) - 1}{x - 1}$ is

If $lim_{x \to \infty} (\frac{x ^{2} + x + 1}{x + 1} - a x - b) = 4$ , then

Let $α (a)$ and $β (a)$ be the roots of the equation $(3 1 + a - 1) x^{2} + (1 + a - 1) x + (6 1 + a - 1) = 0$ where $a > - 1$ . Then $lim_{a \to 0^{+}} α (a)$ and $lim_{a \to 0^{+}} β (a)$ are

If $lim_{x \to 0} [1 + x ln (1 + b^{2})]^{1/ x} = 2 b sin^{2} θ, b > 0$ and $θ \in (- π, π]$ , then the value of $θ$ is

The value of $lim_{x \to 0} \frac{1}{x ^{3}} \int_{0}^{x} \frac{t l n ( 1 + t )}{t ^{4} + 4} d t$ is

Let $L = lim_{x \to 0} \frac{a - a ^{2} - x ^{2} - \frac{x ^{2}}{4}}{x ^{4}}, a > 0$ . If $L$ is finite, then