# Ex 5.1 ,1 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at Jan. 2, 2020 by Teachoo

Last updated at Jan. 2, 2020 by Teachoo

Transcript

Ex 5.1, 1 Prove that the function π (π₯) = 5π₯ β 3 is continuous at π₯ = 0, at π₯ = β3 and at π₯ = 5 Given π(π₯)= 5π₯ β3 At π=π f(x) is continuous at x = 0 if (π₯π’π¦)β¬(π±βπ) π(π) = π(π) limβ¬(xβ0) π(π₯) "= " limβ¬(xβ0) " "(5π₯β3) Putting x = 0 = 5(0) β 3 = β3 π(0) = 5(0) β 3 = 0 β 3 = β3 Since L.H.S = R.H.S β΄ limβ¬(xβ0) π(π₯) = π(0) Hence, f is continuous at π = π At x = β3 f(x) is continuous at x = β3 if ( lim)β¬(xββ3) π(π₯)= π(β3) Since, L.H.S = R.H.S β΄ limβ¬(xββ3) π(π₯) = π(β3) Hence, f is continuous at π =β3 limβ¬(xβ3) π(π₯) "= " limβ¬(xβ3) " "(5π₯β3) Putting x = β3 = 5(β3) β 3 = β18 π(β3) = 5(β3) β 3 = β15 β 3 = β18 Since, L.H.S = R.H.S β΄ limβ¬(xββ3) π(π₯) = π(β3) Hence, f is continuous at π =β3 At π =π f(x) is continuous at x = 5 if ( lim)β¬(xβ5) π(π₯)= π(5) Since, L.H.S = R.H.S β΄ ( lim)β¬(xβ5) π(π₯)= π(5) (πππ)β¬(π₯β5) π(π₯) "= " (πππ)β¬(π₯β5) " "(5π₯β3) Putting x = 5 = 5(5) β 3 = 22 π(5) = 5(5) β 3 = 25 β 3 = 22 Hence, f is continuous at π =π Thus the function is continuous at π =π, at π =βπ & at π =π

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.