Check sibling questions


Transcript

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}

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

About the Author

Davneet Singh

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