Title 5. CONTINUITY & DIFFERENTIABILITY.pdf ✅

TOPIC-1 Continuity Concepts Covered:  Left hand Limit  Right Hand Limit Revision Notes FORMULAE FOR LIMITS: (a) lim cos x x → = 0 1 (b) lim sin x x → x = 0 1 (c) lim tan x x → x = 0 1 (d) lim sin x x → x − = 0 1 1 (e) lim tan x x → x − = 0 1 1 (f) lim log , x x e a x a a → − = > 0 1 0 (g) lim x x e → x − = 0 1 1 (h) lim log ( ) x e x → x + = 0 1 1 (i) lim x a n n x a n x a na → − − − = 1 z For a function f(x), lim x m → f(x) exists if lim x m → − f(x) = lim x m → + f(x). z A function f(x) is continuous at a point x = m if, lim ( ) lim ( ) ( ) xm xm fx fx f m → − → + = = , where lim ( ) x m f x → − is Left Hand Limit of f(x) at x = m and lim ( ) x m f x → + is Right Hand Limit of f(x) at x = m. Also f(m) is the value of function f(x) at x = m. [Board 2020] z A function f(x) is continuous at x = m (say) if, f(m) = lim x m → f(x), i.e., a function is continuous at a point in its domain if the limit value of the function at that point equals the value of the function at the same point. z For a continuous function f(x) atx = m, lim x m → f(x) can be directly obtained by evaluating f(m). z Indeterminate forms or meaningless forms: 0 0 0 1 00 0 , , , , , , . ∞ ∞ × ∞ ∞ − ∞ ∞ ∞ Example-1 Find the value of k for which the function f(x) = sin cos , , x x x x k x − − ≠ =       4 4 4 π π π is continuous at x = x 4 . Sol. lim sin cos x x x x f → − − =       π π π 4 4 4 lim sin cos x x x x k → −       − = π π 4 2 1 2 1 2 4 lim sin cos cos .sin x x x x k → −       − = π π π π 4 2 4 4 4 lim sin x x x k → −       −       = π π π 4 2 4 4 4 2 4 = k k = 1 2 2 UNIT – III : CALCULUS 5 CONTINUITY & DIFFERENTIABILITY CHAPTER Learning Objectives After going through this Chapter, the student would be able to:  Learn the continuity and differentiability of the function.  Relate continuity and differentiability.  Solve problems related to differentiability and continuity. LIST OF TOPICS Topic-1: Continuity Topic-2: Differentiability


Continuity & Differentiability 2. Find the values of p and q, for which: f x x x x p x q x x x ( )= 1 sin 3cos , if < 2 , if = 2 (1 sin ) ( 2 ) , if > 2 3 2 2 - p p - p - p          , is continuous at = 2 x p . A [Delhi Set-1, 2, 3, 2016] Sol. LHL at x=       π 2 = lim ( ) x f x → −π 2 = lim ( sin ) cos x x x → − − π 2 3 2 1 3 = lim x x x → − − π − 2 3 3 2 2 1 3 1 ( sin ) ( sin ) 1⁄2 = lim x x x x x x → − − + + π − + 2 2 1 1 3 1 1 ( sin )( sin sin ) ( sin )( sin ) = lim sin sin sin x x x x → − + + π ( ) + 2 2 1 3 1 1⁄2 = 111 3 1 1 1 2 + + + = ( ) 1⁄2 RHL ( ) at x=π 2 = lim . ( ) ( ) x f x → +π 2 = lim x q x x → + − π π − 2 2 1 2 ( sin ) ( ) Put x = π π 2 2 + → h x . , As h → 0 1⁄2 = q h h lim sin h 0 → − +       − +             1 2 2 2 2 π π π = q h q h h h lim lim h 0 h → → − = 1 4 2 2 4 2 0 2 2 cos sin = q h h q 2 h 2 2 1 0 4 8 2 lim →             × = sin 1⁄2 Also, f π 2       = p Since, f(x) is continuous at x = p 2 \ LHL = RHL = f π 2       \ 1 2 = q p 8 = or q = 4 and p = 1 2 1⁄2 Commonly Made Error Many students commit errors in finding the Left hand limit and Right hand limit. Sufficient time needs to be spent on this topic. Answering Tip 3. For what value of k is the following function continuous at x = − π 6 ? f(x) = 3 6 6 6 sin cos , , x x x x k x + + ≠ − = −        π π π A [SQP 2016-17] Sol. lim ( ) x f x →− π 6 = lim sin cos x x x →− x + + π π 6 3 6 lim ( ) x f x →− π 6 = lim sin cos · x x x →− x +         + π π 6 2 3 2 1 2 6 lim ( ) x f x →− π 6 = lim sin x x →− x +       + π π π 6 2 6 6 1 = 2 1 f −      π 6 = k 1⁄2 For the continuity of f(x) at x = − π 6 , f −      π 6 = lim ( ) x f x →− π 6 or k = 2 1⁄2 [Marking Scheme SQP, 2016-17] (Modified) 4. Find k, if f(x) = k x x x x x x sin ( ), tan sin , π 2 1 0 0 3 + ≤ − >       is continuous at x = 0. R&U [Outside Delhi 2016] This Question is for practice and its solution is given at the end of the chapter.