Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

Transcript

Ex 5.1, 25 Examine the continuity of f, where f is defined by π(π₯)={β(sinβ‘γπ₯βπππ  π₯γ, ππ π₯β 0@&β1, ππ π₯=0)β€ π(π₯)={β(sinβ‘γπ₯βπππ  π₯γ, ππ π₯β 0@&β1, ππ 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¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " 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¬(xβ0^β ) f(x) = limβ¬(hβ0) f(0 β h) = limβ¬(hβ0) f(βh) = limβ¬(hβ0) sin (βh) β cos (βh) = limβ¬(hβ0) (βsin h) β cos h = βsin 0 β cos 0 = 0 β 1 = β1 We know that sin (βx) = β sin x cos (βx) = cos x RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) sin h β cos h = sin 0 β cos 0 = 0 β 1 = β1 And, f(0) = β1 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