Let l n = 2 n + - 2 n 2 n and L n = 2 n + - 2 n 3 n then as n → ∞

lim n → ∞ ⁡ l n does not exist but lim n → ∞ ⁡ L n exists

lim n → ∞ ⁡ l n exists but lim n → ∞ ⁡ L n does not exist

Both the sequences do not have limits

Let l n = 2 n + - 2 n 2 n and L n = 2 n + - 2 n 3 n then as n → ∞

lim n → ∞ ⁡ l n does not exist but lim n → ∞ ⁡ L n exists

Let f x be a polynomial of degree 5 such that x = ± 1 are its critical points. If l i m x → 0 2 + f x x 3 = 4 , then which one of the following is not true?

x = 1 is a point of local minimum and x = - 1 is a point of local maximum

f 1 - 4 f - 1 = 4 . x = 1 is a point of maximum and x = - 1

x = 1 is a point of local maxima of f

Let f x = 1 - x ( 1 + 1 - x ) | 1 - x | cos ⁡ 1 1 - x for x ≠ 1 , then

lim x → 1 - ⁡ f ( x ) does not exist

Determine f defined by f ( x ) = { x 2 sin ( 1 x ) if x ≠ 0 0 if x = 0 is a continuous function.

Given that f ( x ) = { x 2 sin ( 1 x ) if x ≠ 0 0 if x = 0 A function is said to be continuous at x = a , if lim x → a - f x = lim x → a + f x = f a . Now, f 0 = 0 sin 1 0 = 0 Left-hand limit = lim x → 0 − f ( x ) = lim x → 0 − x 2 sin ( 1 x ) = lim h → 0 ( 0 − h ) 2 sin ( 1 0 − h ) = lim h → 0 h 2 sin ( − 1 h ) = - lim h → 0 h sin 1 h 1 h = 0 × 1 = 0 ∵ lim x → 0 sin x x = 1 Right-hand limit = lim x → 0 + f ( x ) = lim x → 0 + x 2 sin ( 1 x ) = lim h → 0 ( 0 + h ) 2 sin ( 1 0 + h ) = lim h → 0 h 2 sin ( 1 h ) = lim h → 0 h sin 1 h 1 h = 0 × 1 = 0 Therefore, lim x → 0 - f x = lim x → 0 + f x = f 0 . And f x is continuous at x = 0

MEDIUM

Earn 100

Let $l_{n} = \frac{2^{n} + {(- 2)}^{n}}{2^{n}}$ and $L_{n} = \frac{2^{n} + {(- 2)}^{n}}{3^{n}}$ then as $n \to \infty$

50% studentsanswered this correctly

Important Questions on Limits and Derivatives

EASY

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Let $f (x)$ be a function defined by $f (x) = \{\begin{array}{cl} 4 x - 5, & if x \leq 2 \\ x - λ, & if x > 2 \end{array}$ . If $\lim_{x \to 2} f (x)$ exists, then the value of $λ$ is

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to 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 and Derivatives>Introduction to Limits

f : [0, 2) \to ℝ

is defined by

f (x) = \{\begin{cases} 1 + \frac{2 x}{k} & for 0 \leq x < 1 \\ k x & for 1 \leq x < 2 \end{cases}

, where

k > 0

and

f

is such that

\lim_{x \to 1^{-}} f (x) = \lim_{x \to 1^{+}} f (x),

then the value of

k^{2}

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

α = \lim_{x \to 0} \frac{x \cdot 2^{x} - x}{1 - \cos x}

and

β = \lim_{x \to 0} \frac{x \cdot 2^{x} - x}{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}},

then

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to 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

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

\underset{x \to 2}{l i m} \frac{3^{x} + 3^{3 - x} - 12}{3^{- \frac{x}{2}} - 3^{1 - x}}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

[x]

is the greatest integer function, then

\lim_{x \to 2^{+}} (\frac{[x]^{3}}{3} - {[\frac{x}{3}]}^{3}) =

EASY

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

The value of the limit

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

is equal to :

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Let

f (x)

be a polynomial of degree

5

such that

x = \pm 1

are its critical points. If

\underset{x \to 0}{l i m} (2 + \frac{f (x)}{x^{3}}) = 4,

then which one of the following is not true?

EASY

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to 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)

EASY

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

The value of

\lim_{x \to 0} \frac{|x|}{x}

is:

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Let

f (x) = \frac{1 - x (1 + |1 - x|)}{| 1 - x |} \cos (\frac{1}{1 - x}) for x \neq 1,

then

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

S^{2} = a t^{2} + 2 bt + c

, then the acceleration is

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

\lim_{x \to 0} | x |^{[\cos x]} = l;

where [.] denotes the greatest integer function, then the value of

l

EASY

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Let

y = f (x) = 2 x^{2} - 3 x + 2

. The differential of

y

when

x

changes from

2

1.99

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

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

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Determine

f

defined by

f (x) = {\begin{array}{cc} x^{2} \sin (\frac{1}{x}) & if x \neq 0 \\ 0 & if x = 0 \end{array}

is a continuous function.

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to 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

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to Limits

Let

F : R \to R

be a function. We say that

f

has:
PROPERTY

1

\lim_{h \to 0} \frac{f (h) - f (0)}{\sqrt{|h|}}

exists and is finite, and
PROPERTY

2

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

exists and is finite.
Then which of the following options is/are correct?

HARD

Mathematics>Differential Calculus>Limits and Derivatives>Introduction to 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

Let ln=2n+-2n2n and Ln=2n+-2n3n then as n→∞

Important Questions on Limits and Derivatives

Learn and Prepare for any exam you want !

Let $l_{n} = \frac{2^{n} + {(- 2)}^{n}}{2^{n}}$ and $L_{n} = \frac{2^{n} + {(- 2)}^{n}}{3^{n}}$ then as $n \to \infty$