Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 8 Find all points of discontinuity of f, where f is defined by π(π₯)={β(|π₯|/π₯, ππ π₯β 0@&0 , 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) 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) Since L.H.L β  R.H.L f(x) is not continuous at x=0 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|)/β = limβ¬(hβ0) β/β = limβ¬(hβ0) 1 = 1 Case 2 : When x < 0 For x < 0, f(x) = (|π₯|)/π₯ f(x) = ((βπ₯))/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) = (|π₯|)/π₯ f(x) = π₯/π₯ f(x) = 1 Since this constant It is continuous β΄ f(x) is continuous for x > 0 Hence, only x = 0is point is discontinuity. β΄ f is continuous for all real numbers except 0. Thus, f is continuous for π β R β {0} (As x > 0, x is positive)