Last updated at Dec. 16, 2024 by Teachoo
Ex 5.1, 2 Examine the continuity of the function f (x) = 2x2 β 1 at x = 3. π(π₯) is continuous at x = 3 if limβ¬(xβ3) π(π₯) = π(3) Since, L.H.S = R.H.S Hence, f is continuous at π =3 (π₯π’π¦)β¬(π±βπ) π(π) "= " limβ¬(xβ3) " "(2π₯2β1) Putting π₯ = 3 = 2(3)2 β 1 = 2 Γ 9 β 1 = 17 π(π) = 2(3)2 β 1 = 2 Γ 9 β 1 = 18β1 = 17
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About the Author
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