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Ex 5.1, 19 - Ex 5.1

Ex 5.1, 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

Ex 5.1, 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1, 19 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 19 (Introduction) Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x. Ex 5.1, 19 Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x. Given g(x) = x − [𝑥] Let c be an integer g(x) is continuous at 𝑥 =𝑐 if L.H.L = R.H.L = 𝑔(𝑐) if lim┬(x→𝑐^− ) 𝑔(𝑥)=lim┬(x→𝑐^+ ) " " 𝑔(𝑥)=𝑔(𝑐) LHL at x → c lim┬(x→𝑐^− ) g(x) = lim┬(h→0) g(c − h) = lim┬(h→0) (𝑐−ℎ)−[𝒄−𝒉] = lim┬(h→0) (𝑐−ℎ)−(𝒄−𝟏) = lim┬(h→0) 𝑐−ℎ−𝑐+1 = lim┬(h→0) −ℎ+1 = 0+1 = 1 RHL at x → c lim┬(x→𝑐^+ ) g(x) = lim┬(h→0) g(c + h) = lim┬(h→0) (𝑐+ℎ)−[𝒄+𝒉] = lim┬(h→0) (𝑐−ℎ)−(𝒄) = lim┬(h→0) −ℎ = 𝟎 Since LHL ≠ RHL ∴ g(x) is not continuous at x = c Hence, g(x) is not continuous at all integral points. Hence proved

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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.