Prove that the function f given by f ( x ) = | x - 3 | , x ∈ R is continuous but not differentiable at x = 3 .

Given, f x = x - 3 , x ∈ R ⇒ f x = 3 - x , x ≤ 3 x - 3 , x ≥ 3 Now let us evaluate the L.H.L of the given function at x = 3 lim x → 3 - f x = lim x → 3 - 3 - x = 3 - 3 = 0 R.H.L at x = 3 lim x → 3 + f x = lim x → 3 + x - 3 = 3 - 3 = 0 And, f ( 3 ) = 0 ∵ lim x → 3 - f x = lim x → 3 + f x = f 3 So, f ( x ) is continuous at x = 3 Now, for Differentiability: L.H.D at x = 3 lim h → 0 f x = lim h → 0 f ( 3 - h ) - f ( 3 ) - h = lim h → 0 3 - ( 3 - h ) - 0 - h = - 1 R.H.D at x = 3 lim h → 0 f x = lim h → 0 f ( 3 + h ) - f ( 3 ) h = lim h → 0 ( 3 + h ) - 3 - 0 h = 1 ∵ L.H.D at x = 3 ≠ R H D at x = 3 Thus, f ( x ) is not differentiable at x = 3 .

Prove that the function f given by f ( x ) = | x - 3 | , x ∈ R is continuous but not differentiable at x = 3 .

Given, f x = x - 3 , x ∈ R ⇒ f x = 3 - x , x ≤ 3 x - 3 , x ≥ 3 Now let us evaluate the L.H.L of the given function at x = 3 lim x → 3 - f x = lim x → 3 - 3 - x = 3 - 3 = 0 R.H.L at x = 3 lim x → 3 + f x = lim x → 3 + x - 3 = 3 - 3 = 0 And, f ( 3 ) = 0 ∵ lim x → 3 - f x = lim x → 3 + f x = f 3 So, f ( x ) is continuous at x = 3 Now, for Differentiability: L.H.D at x = 3 lim h → 0 f x = lim h → 0 f ( 3 - h ) - f ( 3 ) - h = lim h → 0 3 - ( 3 - h ) - 0 - h = - 1 R.H.D at x = 3 lim h → 0 f x = lim h → 0 f ( 3 + h ) - f ( 3 ) h = lim h → 0 ( 3 + h ) - 3 - 0 h = 1 ∵ L.H.D at x = 3 ≠ R H D at x = 3 Thus, f ( x ) is not differentiable at x = 3 .

Show that function f x = ( 7 x + 5 ) , when x ≥ 0 ( 5 - 3 x ) , when x < 0 is a continuous function.

Given, f x = ( 7 x + 5 ) , when x ≥ 0 ( 5 - 3 x ) , when x < 0 We know that a real function f x is continuous at a point x = a of its domain if lim x → a - f x = lim x → a + f x = f a Now let us evaluate the L.H.L of the given function at x = 0 lim x → 0 - f x = lim x → 0 - 5 - 3 x = 5 - 3 × 0 = 5 R.H.L at x = 0 lim x → 0 + f x = lim x → 0 + 7 x + 5 = 7 × 0 + 5 = 5 And, f ( 0 ) = 7 × 0 + 5 = 5 ∵ lim x → 0 - f x = lim x → 0 + f x = f 0 So, f ( x ) is continuous at x = 0 .

Show that function f x = sin x , if x < 0 x , if x ≥ 0 is continuous.

Given f x = sin x , if x < 0 x , if x ≥ 0 We know that a function f x is continuous at a point x = a of its domain if lim x → a - f x = lim x → a + f x = f a L.H.L at x = 0 lim x → 0 - f x = lim x → 0 - sin x = sin 0 o = 0 R.H.L at x = 0 lim x → 0 + f x = lim x → 0 + x = 0 And, f ( 0 ) = 0 ∵ lim x → 0 - f x = lim x → 0 + f x = f 0 So, f ( x ) is continuous at x = 0 .

Show that function f x = x n - 1 x - 1 , when x ≠ 1 n , when x = 1 is continuous.

Given, f x = x n - 1 x - 1 , when x ≠ 1 n , when x = 1 We know that a function f x is continuous at a point x = a of its domain if lim x → a f x = f a Now, lim x → 1 f x = lim x → 1 x n - 1 x - 1 = n 1 n - 1 = n ∵ lim x → a ( x ) n - a n ( x ) - a = n a n - 1 And, f 1 = n Since lim x → 1 f x = f 1 So, f x is continuous at x = 1 .

Show that the function f x = sin x x , when x ≠ 0 2 , when x = 0 is continuous at each point except 0 .

Given, f x = sin x x , when x ≠ 0 2 , when x = 0 We know that a function f x is continuous at a point x = a of its domain if lim x → a f x = f a Now, lim x → 0 f x = lim x → 0 sin x x = 1 And, f ( 0 ) = 2 ∴ lim x → 0 f x ≠ f 0 So, f ( x ) is discontinuous at x = 0 . Now, for x < 0 or x > 0 f x = sin x x We know that trigonometric functions are continuous at all points in its domain. So, sin x is a continuous function and polynomial functions are also continuous functions. So, x is also continuous function. ∴ sin x x is continuous function at each point except x = 0 Hence, the given function is continuous at each point except 0 .

Show that f x = ( 2 x - 1 ) , if x < 2 3 x 2 , if x ≥ 2 is continuous.

Given function f x = ( 2 x - 1 ) , if x < 2 3 x 2 , if x ≥ 2 We know that a function f x is said to be continuous at a point x = a of its domain if lim x → a f x = f a Here, we need to check the continuity of the function at x = 2 Left hand limit at x = 2 lim x → 2 - f x = lim x → 2 - ( 2 x - 1 ) = 2 2 – 1 = 3 Right hand limit at x = 2 lim x → 2 + f x = lim x → 2 + 3 x 2 = 3 ( 2 ) 2 = 3 Also, f ( 2 ) = 3 ( 2 ) 2 = 3 As lim x → 2 - f x = lim x → 2 + f x = f 2 = 3 Therefore, The function f ( x ) is continuous at x = 2 . Hence, the given function is continuous function.