Ex 5.1

Chapter 5 Class 12 Continuity and Differentiability
Serial order wise

### Transcript

Ex 5.1, 24 Determine if f defined by π(π₯)={β( π₯2 sinβ‘γ1/π₯γ, ππ π₯β 0@&0, ππ π₯=0)β€ is a continuous function? 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) = π₯2 sinβ‘γ1/π₯γ Since x2 is continuous and sinβ‘γ1/π₯γ is continuous So, π₯2 sinβ‘γ1/π₯γ 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) (ββ)^2 sinβ‘γ1/((ββ))γ = limβ¬(hβ0) β^2 π = 02 .π = 0 We know that β 1 β€ sin ΞΈ β€ 1 β β 1β€γsin γβ‘γ1/((ββ))γβ€ 1 β΄ γsin γβ‘γ1/((ββ))γ is a finite value Let γsin γβ‘γ1/((ββ))γ = k 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^2 sinβ‘γ1/βγ = limβ¬(hβ0) β^2 π = 02 .π = 0 We know that β 1 β€ sin ΞΈ β€ 1 β β 1β€γsin γβ‘γ1/βγβ€ 1 β΄ γsin γβ‘γ1/βγ is a finite value Let γsin γβ‘γ1/βγ = k And, f(0) = 0 Hence, L.H.L = R.H.L = π(0) β΄ f is continuous at x=0 Hence, π(π) is continuous for all real number

#### Davneet Singh

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.