The limit lim x → 1 1 - cos 2 ( x - 1 ) x - 1 _ _ _ _ _ _ _ _ _

exists and is equal to ( 1 / 2 )

MEDIUM

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The limit $\lim_{x \to 1} \frac{\sqrt{1 - \cos 2 (x - 1)}}{x - 1}_________$

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Important Questions on Limits

HARD

Mathematics>Differential Calculus>Limits>Basics of Limits

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>Basics of Limits

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

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

Let

P = \lim_{x \to 0^{+}} {(1 + \tan^{2} \sqrt{x})}^{\frac{1}{2 x}}

, then

\log P

is equal to

EASY

Mathematics>Differential Calculus>Limits>Basics of Limits

Let

f (x) = \begin{array}{l} x^{2} - 1, 0 < x < 2 \\ 2 x + 3, 2 \leq x < 3 \end{array}

, find the quadratic equation whose roots are

\lim_{x \to 2^{-}} f (x)

and

\lim_{x \to 2^{+}} f (x)

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

Let

x_{n} = {(2^{n} + 3^{n})}^{\frac{1}{2 n}}

for all natural numbers

n

. Then

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

Let

f : R \to R

be a differentiable function satisfying

f^{^{'}} (3) + f^{^{'}} (2) = 0 .

Then

\underset{x \to 0}{l i m} {(\frac{1 + f (3 + x) - f (3)}{1 + f (2 - x) - f (2)})}^{\frac{1}{x}}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

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

is equal to

HARD

Mathematics>Differential Calculus>Limits>Basics of Limits

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

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

For each

x \in R

, let

[x]

be the greatest integer less than or equal to

x

. Then

\underset{x \to 0^{-}}{l i m} \frac{x ([x] + |x|) \sin [x]}{|x|}

is equal to

EASY

Mathematics>Differential Calculus>Limits>Basics of Limits

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

EASY

Mathematics>Differential Calculus>Limits>Basics of Limits

\lim_{x \to 3} \frac{e^{x - 3} - x + 1}{x^{2} - \log (x - 2)}

is equal to

HARD

Mathematics>Differential Calculus>Limits>Basics of Limits

For each

t \in R,

let

[t]

be the greatest integer less than or equal to

t

. Then,

\underset{x \to 1^{+}}{l i m} \frac{(1 - |x| + s i n |1 - x|) s i n ([1 - x] \frac{π}{2})}{|1 - x| [1 - x]}

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

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

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

\lim_{x \to \infty} {(1 + \frac{a}{x} - \frac{4}{x^{2}})}^{2 x} = e^{3}

, then

a

is equal to

HARD

Mathematics>Differential Calculus>Limits>Basics of Limits

Let

[t]

denote the greatest integer

\leq t

. If

λ ε R - \{0, 1\}, \lim_{x \to 0} |\frac{1 - x + |x|}{λ - x + [x]}| = L

, then

L

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

f : R^{+} \to R

be an increasing function such that

f (x) > 0

for all

x

. If

\lim_{x \to \infty} \frac{f (9 x)}{f (3 x)} = 1,

then

\lim_{x \to \infty} \frac{f (6 x)}{f (3 x)} =

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

\lim_{x \to \infty} {(1 + \frac{2}{x})}^{3 x} =

EASY

Mathematics>Differential Calculus>Limits>Basics of Limits

f : R \to (0, \infty)

is an increasing function and if

\lim_{x \to 2018} \frac{f (3 x)}{f (x)} = 1,

then

\lim_{x \to 2018} \frac{f (2 x)}{f (x)}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits>Basics of Limits

\lim_{x \to 0} \frac{e^{x^{2}} - \cos x}{\sin^{2} x}

is equal to

HARD

Mathematics>Differential Calculus>Limits>Basics of Limits

Let $f (x)$ be a polynomial of degree four and having its extreme values at $x = 1$ and $x = 2$ . If $\lim_{x \to 0} [1 + \frac{f (x)}{x^{2}}] = 3$ , then $f (2)$ is equal to

The limit limx→11-cos2(x-1)x-1_________

Important Questions on Limits

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The limit $\lim_{x \to 1} \frac{\sqrt{1 - \cos 2 (x - 1)}}{x - 1}_________$