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

Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 6
Ex 5.1 ,3 - Chapter 5 Class 12 Continuity and Differentiability - Part 7

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Ex 5.1, 3 Examine the following functions for continuity. (c) f (x) = (𝑥^(2 )− 25 )/(𝑥 + 5), x ≠ –5 f (x) = (𝑥^(2 )− 25 )/(𝑥 + 5) Putting x = –5 f (−5) = (〖(−5)〗^(2 )− 25 )/(−5 + 5) = (25− 25 )/(−5 + 5) = 0/0 = Undefined Hence, f(x) is not defined at x = −5 So, we check for continuity at all points except −5 Let c be any real number except −5. f is continuous at 𝑥 = 𝑐 if (𝐥𝐢𝐦)┬(𝐱→𝒄) 𝒇(𝒙) = 𝒇(𝒄) LHS (𝐥𝐢𝐦)┬(𝐱→𝒄) 𝒇(𝒙) = lim┬(x→𝑐) (𝑥^2− 25)/(𝑥 + 5) = lim┬(x→𝑐) ((𝑥 − 5) (𝑥 + 5))/(𝑥 + 5) = lim┬(x→𝑐) 𝑥−5 Putting x = c = c − 5 RHS f (c) = (𝑐^(2 )− 25 )/(𝑐 + 5) = ((𝑐 − 5)(𝑐 + 5))/((𝑐 + 5)) = c − 5 Let c be any real number except −5. f is continuous at 𝑥 = 𝑐 if (𝐥𝐢𝐦)┬(𝐱→𝒄) 𝒇(𝒙) = 𝒇(𝒄) Since, L.H.S = R.H.S ∴ Function is continuous at x = c (except −5) Thus, we can write that f is continuous for all real numbers except −5 ∴ f is continuous at each 𝐱 ∈ R − {−𝟓}

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.