Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 9 Find all points of discontinuity of f, where f is defined by π(π₯)={β(π₯/|π₯| , ππ π₯<0@&β1 , ππ π₯β₯ 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) Since there are two different functions on the left & right of 0, we take LHL & RHL . if limβ¬(xβ0^β ) π(π₯)=limβ¬(xβ0^+ ) " " π(π₯)= π(0) And, f(0) = β1 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β)/|ββ| = limβ¬(hβ0) (ββ)/β = limβ¬(hβ0) β1 = β1 RHL at x β 0 limβ¬(xβ0^+ ) f(x) = limβ¬(hβ0) f(0 + h) = limβ¬(hβ0) f(h) = limβ¬(hβ0) β1 = β1 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x = β3 Case 2 : When x < 0 For x < 0, f(x) = π₯/(|π₯|) f(x) = π₯/((βπ₯)) f(x) = β1 Since this constant It is continuous β΄ f(x) is continuous for x < 0 (As x < 0, x is negative) Case 3 : When x > 0 For x > 0, f(x) = β1 Since this constant It is continuous β΄ f(x) is continuous for x > 0 β΄ f is continuous for all real numbers Thus, f is continuous for πβ R