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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.1, 8 Find all points of discontinuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(|π‘₯|/π‘₯, 𝑖𝑓 π‘₯β‰ 0@&0 , 𝑖𝑓 π‘₯=0)─ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x > 0 When x < 0 Case 1 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since L.H.L β‰  R.H.L f(x) is not continuous at x=0 LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) (|βˆ’β„Ž|)/(βˆ’β„Ž) = lim┬(hβ†’0) β„Ž/(βˆ’β„Ž) = lim┬(hβ†’0) βˆ’1 = βˆ’1 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) (|β„Ž|)/β„Ž = lim┬(hβ†’0) β„Ž/β„Ž = lim┬(hβ†’0) 1 = 1 Case 2 : When x < 0 For x < 0, f(x) = (|π‘₯|)/π‘₯ f(x) = ((βˆ’π‘₯))/π‘₯ f(x) = βˆ’1 Since this constant It is continuous ∴ f(x) is continuous for x < 0 (As x < 0, x is negative) Case 3 : When x > 0 For x > 0, f(x) = (|π‘₯|)/π‘₯ f(x) = π‘₯/π‘₯ f(x) = 1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Hence, only x=0 is point is discontinuity. ∴ f is continuous for all real numbers except 0. Thus, f is continuous for π‘₯ ∈ R βˆ’ {0} (As x > 0, x is positive)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.