Define a sequence S n of real numbers by, S n = ∑ k = 0 n 1 n 2 + k for n ≥ 1 . Then lim n → ∞ S n

exists and lies in the interval [ 1 , 2 )

exists and lies in the interval ( 0 , 1 )

exists and lies in the interval [ 2 , ∞ )

MEDIUM

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$\lim_{n \to \infty} \{(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) . . . . (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})\}$ is equal to

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Important Questions on Limits

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

Let

a = m i n \{x^{2} + 2 x + 3 : x \in R\}

and

b = \lim_{θ \to 0} \frac{1 - \cos θ}{θ^{2}} .

Then

\sum_{r = 0}^{n} a^{r} b^{n - r}

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

The value of

\lim_{x \to 0} \frac{(1 - \cos 2 x) (3 + \cos x)}{x \tan 4 x}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{n \to \infty} \sin [π \sqrt{n^{2} + 1}]

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

The value of

\lim_{h \to 0} \{\frac{\sqrt{3} \sin (\frac{π}{6} + h) - \cos (\frac{π}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)}\}

is :

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

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)}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

Define a sequence

\{S_{n}\}

of real numbers by,

S_{n} = \sum_{k = 0}^{n} \frac{1}{\sqrt{n^{2} + k}} for n \geq 1

. Then

\lim_{n \to \infty} S_{n}

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

f (x) = |\begin{array}{l} \sin x & \cos x & \tan x \\ x^{3} & x^{2} & x \\ 2 x & 1 & x \end{array}|,

x \in (- \frac{π}{2}, \frac{π}{2}),

then

\lim_{x \to 0} \frac{f (x)}{x^{2}}

is equal to

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{x \to 0} \frac{x^{a} \sin^{b} x}{\sin x^{c}}, a, b, c, \in R ~ \{0\}

exists and has non-zero value, then

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{x \to 0} \frac{\sin ({πcos}^{2} x)}{x^{2}}

is equal to

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

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}])

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

The value of

\lim_{n \to \infty} \frac{[r] + [2 r] + . . . + [n r]}{n^{2}},

where

r

is non-zero real number and

[r]

denotes the greatest integer less than or equal to

r,

is equal to :

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

[\cdot]

denote the greatest integer function then

\lim_{n \to \infty} \frac{[x] + [2 x] + \dots . + [n x]}{n^{2}}

is -

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

The value of the limit

\lim_{θ \to 0} \frac{\tan (π \cos^{2} θ)}{\sin (2 π \sin^{2} θ)}

is equal to :

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

Let

\lim_{x \to 0} \frac{x 5^{x} - x}{1 - \cos x}

is equal to

HARD

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

[\cdot]

denotes the greatest integer function then

\lim_{n \to \infty} \frac{[x] + [2 x] + \dots + [n x]}{n^{2}}

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{x \to 0} \frac{\sin^{2} (π \cos^{4} x)}{x^{4}}

is equal to :

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{θ \to \frac{π}{4}} \frac{\sqrt{2} - \cos θ - \sin θ}{(4 θ - π)^{2}}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

α

is the positive root of the equation,

p (x) = x^{2} - x - 2 = 0

, then

\lim_{x \to α^{+}} \frac{\sqrt{1 - \cos p (x)}}{x + α - 4}

is equal to

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

f (x) = \sqrt{\frac{x - \sin x}{x + \cos^{2} x}}

,

then

\lim_{x \to \infty} f (x)

is equal to

EASY

Mathematics>Differential Calculus>Limits>Limits of Trigonometric Functions

\lim_{x \to 0} \frac{1 - \cos m x}{1 - \cos n x} =

limn→∞212-213212-215....212-212n+1 is equal to

Important Questions on Limits

Learn and Prepare for any exam you want !

$\lim_{n \to \infty} \{(2^{\frac{1}{2}} - 2^{\frac{1}{3}}) (2^{\frac{1}{2}} - 2^{\frac{1}{5}}) . . . . (2^{\frac{1}{2}} - 2^{\frac{1}{2 n + 1}})\}$ is equal to