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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 4 Show that the function f given by 𝑓(π‘₯)={β–ˆ(π‘₯3+3, 𝑖𝑓 π‘₯β‰ 0@1, 𝑖𝑓 π‘₯=0)─ is not continuous at x = 0. f(x) is continuous at π‘₯ =0 if L.H.L = R.H.L = 𝑓(0) If (π‘™π‘–π‘š)┬(π‘₯β†’0^βˆ’ ) 𝑓(π‘₯)=(π‘™π‘–π‘š)┬(π‘₯β†’0^+ ) " " 𝑓(π‘₯)= 𝑓(0) Finding LHL and 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) (βˆ’h)3 + 3 = 03 + 3 = 3 RHL at x β†’ 0 lim┬(xβ†’0^+ ) f(x) = lim┬(hβ†’0) f(0 + h) = lim┬(hβ†’0) f(h) = lim┬(hβ†’0) h3 + 3 = 03 + 3 = 3 But, f(0) = 1 So, LHL = RHL β‰  f(0) Hence, f is not continuous at 𝒙 = 𝟎

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.