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Chapter 5 Class 12 Continuity and Differentiability
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Example 13 Discuss the continuity of the function f given by π(π₯)={β(& π₯, ππ π₯β₯[email protected]& π₯2 , ππ π₯<0)β€ π(π₯)={β(& π₯, ππ π₯β₯[email protected]& π₯2 , 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 = 0 When x < 0 When x > 0 Case 1 : 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) (βh)2 = (β0)2 = 0 RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) h = 0 & π(0) = π₯ = 0 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = 0 Case 2 : When x < 0 For x < 0, f(x) = π₯^2 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 0 Case 3 : When x > 0 For x > 0, f(x) = x Since this a polynomial It is continuous β΄ f(x) is continuous for x > 0 Hence, there is no point of discontinuity Thus, f is continuous for all πβπ