Discuss the continuity of the function f x = lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n at x = 1 .

Given f x = lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n Case 1 : When x < 1 We have lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n = log 2 + x - 0 1 + 0 = log 2 + x Case 2 : When x > 1 We have lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n = log 2 + x x 2 n - sin x 1 + 1 x 2 n = 0 - sin x 1 + 0 = - sin x Thus, f x is discontinuous at x = 1 .

Discuss the continuity of the function f x = lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n at x = 1 .

Given f x = lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n Case 1 : When x < 1 We have lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n = log 2 + x - 0 1 + 0 = log 2 + x Case 2 : When x > 1 We have lim n → ∞ ln 2 + x - x 2 n sin x 1 + x 2 n = log 2 + x x 2 n - sin x 1 + 1 x 2 n = 0 - sin x 1 + 0 = - sin x Thus, f x is discontinuous at x = 1 .

Consider the function g x = 1 − a x + x a x ln a a x x 2 for x < 0 2 x a x − x ln 2 − x ln a − 1 x 2 for x > 0 where a > 0 . Find the value of ' a ' & ' g 0 ' so that the function g x is continuous at x = 0 .

Given g x = 1 − a x + x a x ln a a x x 2 for x < 0 2 x a x − x ln 2 − x ln a − 1 x 2 for x > 0 L . H . L = lim x → 0 - 1 − a x + x a x ln a a x x 2 = lim x → 0 - 1 - 1 + x ln a 1 ! + x 2 ln a 2 2 ! + x ln a 1 + x ln a 1 ! a x x 2 = lim x → 0 - 1 2 x ln a 2 a x · x 2 L . H . L = 1 2 ln a 2 … 1 R . H . L = lim x → 0 + 1 + x ln 2 a 1 ! + x ln 2 a 2 2 ! - x ln 2 a - 1 x 2 = 1 2 ln 2 a 2 … 2 Now, L . H . L = R . H . L 1 2 ln a 2 = 1 2 ln 2 a 2 ⇒ a = 1 2 g 0 = 1 2 ln 2 2 = 1 4 · 1 2 ln 2 2 = 1 8 ln 2 2

Let f ( x ) = π 2 − sin − 1 1 − x 2 ⋅ sin − 1 1 − x 2 x − x 3 for x ≠ 0 π 2 for x = 0 where x is the fractional part of x . Consider another function g x ; such that g x = f x for x ≥ 0 2 2 f x for x < 0 Discuss the continuity of the functions f x and g x at x = 0 .

Given f x = π 2 − sin − 1 1 − x 2 ⋅ sin − 1 1 − x 2 x − x 3 for x ≠ 0 π 2 for x = 0 lim x → 0 + f x = lim x → h π 2 − sin − 1 1 − h 2 ⋅ sin − 1 1 − h 2 h − h 3 lim x → 0 + f x = lim x → h π 2 − sin − 1 1 − h 2 ⋅ sin − 1 1 − h 2 h − h 3 lim x → 0 + f x = lim x → h cos - 1 1 - h 2 2 1 - h 2 × sin - 1 1 - h h = π 2 lim x → 0 - f x = lim x → h π 2 - sin - 1 1 - 1 - h 2 sin - 1 1 - 1 - h 2 1 - h - 1 - h 3 lim x → 0 - f x = lim x → h π 2 sin - 1 h 2 1 - h 2 - h h = π 4 2 So, f x is discontinuous at x = 0 . Now, g x = π 2 x ≥ 0 2 2 × π 4 2 x < 0 g x = π 2 x ≥ 0 π 2 x < 0 So, g x is continuous at x = 0 .

f x = a sin x - a tan x tan x - sin x for x > 0 f x = ln 1 + x + x 2 + ln 1 - x + x 2 sec x - cos x for x < 0 , if ' f ' is continuous at x = 0 , find ' a ' now if g x = ln 2 - x a · cot x - a for x ≠ a , a ≠ 0 , a > 0 . If ' g ' is continuous at x = a then show that g e - 1 = - e .

Given f x = a sin x - a tan x tan x - sin x ⇒ lim x → 0 + f x = lim x → 0 + a sin x - a tan x tan x - sin x = a tan x a sin x - tan x - sin x - tan x = - ln a Also f x = ln 1 + x + x 2 + ln 1 - x + x 2 sec x - cos x ⇒ lim x → 0 - f x = lim x → 0 - ln 1 + x 2 + x 4 x 2 + x 4 × x 2 1 - cos x × 1 + x 2 1 + cos x = 1 For f ( x ) to be continuous ln a = - 1 ⇒ a = e - 1 Given g x = ln 2 - x a · cot x - a ⇒ lim x → a g x = lim x → a ln 1 + a - x a a - x × cos x - a sin x - a × x - a - a ⇒ lim x → a g x = - 1 a g a = - 1 a Therefore, g e - 1 = - 1 e - 1 = - e

Let x denote the greatest integer function & f x be defined in a neighbourhood of 2 by f x = exp x + 2 ln 4 x + 1 4 − 16 4 x − 16 , x < 2 A 1 − cos x − 2 x − 2 tan x − 2 , x > 2 Find the value of A & f 2 in order that f x may be continuous at x = 2 .

We have, f x = exp x + 2 ln 4 x + 1 4 − 16 4 x − 16 , x < 2 A 1 − cos x − 2 x − 2 tan x − 2 , x > 2 ⇒ f x = 4 x + 2 x + 1 4 − 16 4 x − 16 , x < 2 A 1 − cos x − 2 x − 2 tan x − 2 , x > 2 Now, lim x → 2 - f x = lim h → 0 f 2 - h = lim h → 0 4 2 - h + 2 2 - h + 1 4 − 16 4 2 - h − 16 = lim h → 0 f 2 - h = lim h → 0 2 4 - h 3 - h 2 − 4 2 2 2 - h 2 − 4 2 = lim h → 0 2 4 · 2 - h - 2 4 2 2 - h + 4 2 2 - h - 4 = lim h → 0 2 4 2 - h - 1 4 · 4 · 2 - h + 1 2 - h - 1 = 2 4 8 · 4 = 1 2 And, lim x → 2 - f x = lim h → 0 f 2 + h = lim h → 0 A 1 − cos h h tan h = lim h → 0 A 2 sin 2 h 2 h 2 tan h h = lim h → 0 A sin 2 h 2 2 h 2 4 tan h h = A 2 lim h → 0 sin h 2 h 2 2 tan h h = A 2 Since, lim x → 0 sin x x = 1 and lim x → 0 tan x x = 1 . Since, f x is continuous at x = 2 , therefore A 2 = 1 2 ⇒ A = 1 And hence for continuity f 2 = 1 2

Let f x = = sin x + cos x cosec x ; π 2 < x < 0 a ; x = 0 e 1 x + e 2 x + e 3 x a e 2 x + b e 3 x ; 0 < x < π 2 . If f x is continuous at x = 0 , find the value of a 2 + b 2 .

We have, = sin x + cos x cosec x ; π 2 < x < 0 a ; x = 0 e 1 x + e 2 x + e 3 x a e 2 x + b e 3 x ; 0 < x < π 2 Now, lim x → 0 - f x = lim h → 0 - sin h + cos h - cosec h = e lim h → 0 - sin h + cos h - 1 - sin h = e lim h → 0 - 2 sin h 2 cos h 2 - 2 sin 2 h 2 - 2 sin h 2 cos h 2 = e lim h → 0 cos h 2 + sin h 2 cos h 2 = e lim h → 0 1 + tan h 2 = e And, lim x → 0 + f x = lim h → 0 e 1 h + e 2 h + e 3 h a e 2 h + b e 3 h = lim h → 0 e - 2 h + e - 1 h + 1 a e - 1 h + b = 1 b Since, f x is continuous at x = 0 , so lim x → 0 + f x = lim x → 0 - f x = f 0 ∴ a = e , b = 1 e Hence, a 2 + b 2 = e 2 + e - 2

Let y n x = x 2 + x 2 1 + x 2 + x 2 1 + x 2 2 + … x 2 1 + x 2 n - 1 and y x = lim n → ∞ y n x . Discuss the continuity of y n x n = 1 , 2 , 3 … n and y x at x = 0

Given y n x = x 2 + x 2 1 + x 2 + x 2 1 + x 2 2 + … x 2 1 + x 2 n - 1 ⇒ y n x = x 2 1 - 1 1 + x 2 n 1 - 1 1 + x 2 ⇒ y n x = 1 + x 2 n - 1 1 + x 2 n - 1 y x = lim n → ∞ y n x = lim n → ∞ 1 + x 2 n - 1 1 + x 2 n - 1 = lim n → ∞ 1 + x 2 n - 1 1 + x 2 - 1 1 + x 2 n - 1 1 + x 2 n - 1 ⇒ y x = lim n → ∞ y n x = 1 + x 2 - 1 1 + x 2 ∞ ⇒ y x = 1 + x 2 , y ( 0 ) = 1 As y n x = 1 + x 2 n - 1 1 + x 2 n - 1 then when n = 1 , ⇒ y 1 x = 1 + x 2 1 - 1 1 + x 2 0 = x 2 When n = 2 , we get y 2 x = x 4 + 2 x 2 x 2 + 1 Hence y n x = 1 + x 2 1 - 1 1 + x 2 n if x ≠ 0 0 if x = 0 Hence y n x is continuous at x = 0 ∀ n ∈ N . Again y x = 1 + x 2 if x ≠ 0 0 if x = 0 Hence y x is discontinuous at x = 0 .

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Discuss the continuity of the function $f (x) = \lim_{n \to \infty} \frac{\ln (2 + x) - x^{2 n} \sin x}{1 + x^{2 n}}$ at $x = 1$ .

Important Questions on Continuity and Differentiability

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 5

Consider the function $g (x) = [\begin{matrix} \frac{1 - a^{x} + x a^{x} \ln a}{a^{x} x^{2}} & for x < 0 \\ \frac{2^{x} a^{x} - x \ln 2 - x \ln a - 1}{x^{2}} & for x > 0 \end{matrix}$ where $a > 0$ .

Find the value of $' a' &' g (0)'$ so that the function $g (x)$ is continuous at $x = 0$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 6

Let $f (x) = [\begin{matrix} \frac{(\frac{π}{2} - \sin^{- 1} (1 - {|x|}^{2})) \cdot \sin^{- 1} (1 - (x))}{\sqrt{2} (\{x\} - {\{x\}}^{3})} & for x \neq 0 \\ \frac{π}{2} & for x = 0 \end{matrix}$ where $\{x\}$ is the fractional part of $x$ . Consider another function $g (x)$ ; such that

$g (x) = [\begin{matrix} f (x) & for x \geq 0 \\ 2 \sqrt{2} f (x) & for x < 0 \end{matrix}$

Discuss the continuity of the functions $f (x)$ and $g (x)$ at $x = 0$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 7

$f (x) = \frac{a^{\sin x} - a^{\tan x}}{\tan x - \sin x} for x > 0$

$f (x) = \frac{\ln (1 + x + x^{2}) + \ln (1 - x + x^{2})}{\sec x - \cos x}$ for $x < 0$ , if $' f'$ is continuous at $x = 0$ , find $' a'$ now if $g (x) = \ln (2 - \frac{x}{a}) \cdot \cot (x - a)$ for $x \neq a, a \neq 0, a > 0$ . If $' g'$ is continuous at $x = a$ then show that $g (e^{- 1}) = - e$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 8

Let $[x]$ denote the greatest integer function & $f (x)$ be defined in a neighbourhood of $2$ by $f (x) = [\begin{matrix} \{\frac{{(\exp \{(x + 2) \ln 4\})}^{\frac{[x + 1]}{4}} - 16}{4^{x} - 16}\}, & x < 2 \\ A \{\frac{1 - \cos (x - 2)}{(x - 2) \tan (x - 2)}\}, & x > 2 \end{matrix}$

Find the value of $A$ & $f (2)$ in order that $f (x)$ may be continuous at $x = 2$ .

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 9

Let $f (x) = \{\begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$ Determine the form of $g (x) = f [f (x)]$ and hence find the points of discontinuity of $g$ , if any

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 10

Let

f (x) = = \{\begin{cases} \begin{matrix} {(\sin x + \cos x)}^{cosec x} & ; & \frac{π}{2} < x < 0 \end{matrix} \\ a; x = 0 \\ \begin{matrix} \frac{e^{\frac{1}{x}} + e^{\frac{2}{x}} + e^{\frac{3}{|x|}}}{a e^{\frac{2}{x}} + b e^{\frac{3}{|x|}}} & ; & 0 < x < \frac{π}{2} \end{matrix} \end{cases}

. If

f (x)

is continuous at

x = 0

, find the value of

(a^{2} + b^{2})

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 12

Let

y_{n} (x) = x^{2} + \frac{x^{2}}{1 + x^{2}} + \frac{x^{2}}{{(1 + x^{2})}^{2}} + \dots \frac{x^{2}}{{(1 + x^{2})}^{n - 1}}

and

y (x) = \lim_{n \to \infty} y_{n} (x)

. Discuss the continuity of

y_{n} (x) (n = 1, 2, 3 \dots n)

and

y (x)

x = 0

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 27 - Continuity and Differentiability>EXERCISE-4>Q 14

Examine the continuity at $x = 0$ of the sum function of the infinite series:

$\frac{x}{x + 1} + \frac{x}{(x + 1) (2 x + 1)} + \frac{x}{(2 x + 1) (3 x + 1)} + \dots \infty$

Discuss the continuity of the function fx=limn→∞ln2+x-x2nsinx1+x2n at x=1.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Discuss the continuity of the function $f (x) = \lim_{n \to \infty} \frac{\ln (2 + x) - x^{2 n} \sin x}{1 + x^{2 n}}$ at $x = 1$ .