Ex 5.1, 23 - Find all points of discontinuity of f(x) = {sinx/x

Ex 5.1, 23 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 23 - Chapter 5 Class 12 Continuity and Differentiability - Part 3
Ex 5.1, 23 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 23 Find all points of discontinuity of f, where 𝑓(π‘₯)={β–ˆ(sin⁑π‘₯/π‘₯, 𝑖𝑓 π‘₯<0@&π‘₯+1, 𝑖𝑓 π‘₯β‰₯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 For x < 0, f(x) = sin⁑π‘₯/π‘₯ Since sin x and x are continuous So, sin⁑π‘₯/π‘₯ is continuous ∴ f(x) is continuous for x < 0 Case 2 : When x = 0 f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . 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) sin⁑〖 (βˆ’β„Ž)γ€—/((βˆ’β„Ž)) = 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+1 = 0 + 1 = 1 & 𝑓(0) = π‘₯+1 = 0+1 = 1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x = 0 Case 3 : When x > 0 For x > 0, f(x) = x + 1 Since this a polynomial It is continuous ∴ f(x) is continuous for x > 1 Hence, there is no point of discontinuity

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo