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Ex 5.1, 8 - Find points of discontinuity f(x) = {|x|/x, if x=0

Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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 = 0is point is discontinuity. ∴ f is continuous for all real numbers except 0. Thus, f is continuous for 𝒙 ∈ R βˆ’ {0} (As x > 0, x is positive)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.