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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 x is 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 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 numbers

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

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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 and Computer Science at Teachoo