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Ex 5.1, 25 - Examine continuity of f(x) = {sin x - cos x, -1

Ex 5.1, 25 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1, 25 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1, 25 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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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‰ [email protected]&βˆ’1, 𝑖𝑓 π‘₯=0)─ 𝑓(π‘₯)={β–ˆ(sin⁑〖π‘₯βˆ’π‘π‘œπ‘  π‘₯γ€—, 𝑖𝑓 π‘₯β‰ [email protected]&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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.