Last updated at Dec. 16, 2024 by Teachoo
Ex 5.1, 25 Examine the continuity of f, where f is defined by π(π₯)={β(sinβ‘γπ₯βπππ π₯γ, ππ π₯β 0@&β1, ππ π₯=0)β€ π(π₯)={β(sinβ‘γπ₯βπππ π₯γ, ππ π₯β 0@&β1, ππ π₯=0)β€ Since we need to find continuity at of the function We check continuity for different values of x When x β 0 When x = 0 Case 1 : When x β 0 For x β 0, f(x) = sin x β cos x Since sin x is continuous and cos x is continuous So, sin x β cos x is continuous β΄ f(x) is continuous for x β 0 Case 2 : When x = 0 f(x) is continuous at π₯ =0 if L.H.L = R.H.L = π(0) if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β 0 limβ¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) sin (βh) β cos (βh) = limβ¬(hβ0) (βsin h) β cos h = βsin 0 β cos 0 = 0 β 1 = β1 We know that sin (βx) = β sin x cos (βx) = cos x RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) sin h β cos h = sin 0 β cos 0 = 0 β 1 = β1 And, f(0) = β1 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = 0 Hence, π(π) is continuous for all real numbers
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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