Last updated at March 10, 2021 by Teachoo

Transcript

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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.