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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise

Transcript

Ex 5.1, 25 Examine the continuity of f, where f is defined by 𝑓(π‘₯)={β–ˆ(sin⁑〖π‘₯βˆ’π‘π‘œπ‘  π‘₯γ€—, 𝑖𝑓 π‘₯β‰ 0@&βˆ’1, 𝑖𝑓 π‘₯=0)─ 𝑓(π‘₯)={β–ˆ(sin⁑〖π‘₯βˆ’π‘π‘œπ‘  π‘₯γ€—, 𝑖𝑓 π‘₯β‰ 0@&βˆ’1, 𝑖𝑓 π‘₯=0)─ 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 – cos x Since sin x is continuous and cos x is continuous So, sin x – cos xis continuous ∴ 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 = 𝑓(0) if lim┬(xβ†’0^βˆ’ ) 𝑓(π‘₯)=lim┬(xβ†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . LHL at x β†’ 0 lim┬(xβ†’0^βˆ’ ) f(x) = lim┬(hβ†’0) f(0 βˆ’ h) = lim┬(hβ†’0) f(βˆ’h) = lim┬(hβ†’0) sin (βˆ’h) βˆ’ cos (βˆ’h) = lim┬(hβ†’0) (βˆ’sin h) βˆ’ cos h = βˆ’sin 0 βˆ’ cos 0 = 0 βˆ’ 1 = βˆ’1 We know that sin (βˆ’x) = βˆ’ sin x And cos (βˆ’x) = cos x RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) sin h βˆ’ cos h = sin 0 βˆ’ cos 0 = 0 βˆ’ 1 = βˆ’1 And, f(0) = βˆ’1 Hence, L.H.L = R.H.L = 𝑓(0) ∴ f is continuous at x=0 Hence, 𝒇(𝒙) is continuous for all real value of 𝒙 .

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.