Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Limits - Limit exists
Last updated at May 29, 2023 by Teachoo
Ex 12.1, 31 If the function f(x) satisfies limβ¬(x β 1) (π(π₯) β 2)/(π₯2 β 1) = Ο , evaluate limβ¬(xβ1) f(x) . Given limβ¬(xβ1) (π(π₯) β 2)/(π₯^2 β 1) = Ο (limβ¬(xβ1) π(π₯) β 2)/(limβ¬(xβ1) γ(π₯γ^2 β 1) ) = Ο limβ¬(xβ1) (f(x) β 2) = Ο Γ limβ¬(xβ1) (x2 β 1) limβ¬(xβ1) f(x) β limβ¬(xβ1) 2 = Ο (limβ¬(xβ1) x2 β limβ¬(xβ1) 1) By Algebra of limits (πππ)β¬(π₯βπ) (π(π₯))/(π(π₯)) = ((πππ)β¬(π₯βπ) π(π₯))/((πππ)β¬(π₯βπ) π(π₯)) (limβ¬(xβ1) π(π₯) β 2)/(limβ¬(xβ1) γ(π₯γ^2 β 1) ) = Ο limβ¬(xβ1) (f(x) β 2) = Ο Γ limβ¬(xβ1) (x2 β 1) limβ¬(xβ1) f(x) β limβ¬(xβ1) 2 = Ο (limβ¬(xβ1) x2 β limβ¬(xβ1) 1) Finding limits, putting x = 1 limβ¬(xβ1) f(x) β 2 = Ο Γ ((1)2 β 1) limβ¬(xβ1) f(x) β 2 = Ο Γ 0 limβ¬(xβ1) f(x) β 2 = Ο Γ 0 limβ¬(xβ1) f(x) β 2 = 0 limβ¬(xβ1) f(x) = 2 Thus (πππ)β¬(π±βπ) f (x) = 2