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Last updated at Nov. 30, 2019 by Teachoo
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Ex 13.1, 24 (Method 1) Find (πππ)β¬(π₯β1) f(x), where f(x) = {β(π₯2 β1,@βπ₯2 β1,)β€ β 8(π₯ β€1@π₯>1) The Limit at x = 1 will be (πππ)β¬(π₯β1) f(x) = limβ¬(γxβ1γ^β ) f(x) =(πππ)β¬(γπ₯β1γ^+ ) f(x) (πππ)β¬(γπβπγ^β ) f(x) = (πππ)β¬(π₯β1) x2 β 1 = (1)2 β 1 = 1 β 1 = 0 (πππ)β¬(γπβπγ^+ ) f(x) = (πππ)β¬(π₯β1) (βx2 β 1) = β(1)2 β 1 = β1 β 1 = β2 Thus, (πππ)β¬(γπβπγ^+ )f(x) β (πππ)β¬(γπβπγ^β )f(x) Since, Left Hand Limit & Right Hand Limit are not equal Hence (π₯ππ)β¬(πβπ) f(x) does not exit Ex 13.1, 24 (Method 2) Find (πππ)β¬(π₯β1) f(x), where f(x) = {β(π₯2 β1,@βπ₯2 β1,)β€ β 8(π₯ β€1@π₯>1) The Limit at x = 1 will be (πππ)β¬(π₯β1) f(x) = limβ¬(γxβ1γ^β ) f(x) =(πππ)β¬(γπ₯β1γ^+ ) f(x) LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) (1 β h)2 β1 = (1 β 0)2 β 1 = (1)2 β 1 = 1 β 1 = 0 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) β(1 + h)2 β 1 = β(1 + 0)2 β 1 = β(1)2 β 1 = β1 β 1 = β2 Since LHL β RHL β΄ (πππ)β¬(πβπ) f(x) doesnβt exist
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