Examine the continuity of the function: f x = | x | cos 1 x , x ≠ 0 0 , x = 0 at the point x = 0 .

Given function: f x = | x | cos 1 x , x ≠ 0 0 , x = 0 Now, check continuity at x = 0 , So, LHL = lim x → 0 - f x = lim h → 0 0 - h cos 1 0 - h = lim h → 0 h cos 1 h = 0 × an oscillating number between - 1 and 1 = 0 , RHL = lim x → 0 + f x = lim h → 0 0 + h cos 1 0 + h = lim h → 0 h cos 1 h = 0 × an oscillating number between - 1 and 1 = 0 And f 0 = 0 Thus, LHL = RHL = f 0 So, f x is continuous at x = 0 .

Examine the continuity of the function: f x = | x - a | sin 1 x - a , x ≠ a 0 , x = a at the point x = a .

Given function: f x = | x - a | sin 1 x - a , x ≠ a 0 , x = a Now, check continuity at x = a , So, LHL = lim x → a - f x = lim h → 0 a - h - a sin 1 a - h - a = lim h → 0 h sin 1 h = 0 × [an oscillating number between - 1 and 1 ] = 0 , RHL = lim x → a + f x = lim h → 0 a + h - a sin 1 a + h - a = lim h → 0 h sin 1 h = 0 × [ an oscillating number between - 1 and 1 ] = 0 And f a = 0 Thus, LHL = RHL = f a So, f x is continuous at x = a .

Examine the continuity of the function: f x = e x - 1 log ( 1 + 2 x ) , x ≠ 0 7 , x = 0 at the point x = 0 .

Given function: f x = e x - 1 log ( 1 + 2 x ) , x ≠ 0 7 , x = 0 We know that, a function is continuous at x = a , if lim x → a f x = f a . Given that, f 0 = 7 Therefore, lim x → 0 f x = lim x → 0 e x - 1 log ( 1 + 2 x ) = lim x → 0 e x - 1 x log ( 1 + 2 x ) x [As x → 0 , therefore x ≠ 0 ] = lim x → 0 e x - 1 x lim x → 0 log ( 1 + 2 x ) x = lim x → 0 e x - 1 x lim x → 0 log ( 1 + 2 x ) 2 x × 2 = lim x → 0 e x - 1 x 2 lim 2 x → 0 log ( 1 + 2 x ) 2 x = 1 2 [Using lim x → 0 e x - 1 x = 1 , lim f x → 0 log 1 + f x f x = 1 ] So, clearly lim x → 0 f x ≠ f ( 0 ) i.e., limit exists and function also defined but they are not equal. Thus, the given function is discontinuous at x = 0 .

Examine the continuity of the function: f x = sin x x + cos x , x ≠ 0 2 , x = 0 at the point x = 0 .

Given function: f x = sin x x + cos x , x ≠ 0 2 , x = 0 We know that, a function is continuous at x = a , if lim x → a f x = f a . Given that, f 0 = 2 Therefore, lim x → 0 f x = lim x → 0 sin x x + cos x = lim x → 0 sin x x + lim x → 0 cos x = 1 + cos 0 [Using lim θ → 0 sin a θ θ = a ] = 1 + 1 = 2 So, clearly, lim x → 0 f x = f ( 0 ) Thus, the given function is continuous at x = 0 .