Example 4 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at August 2, 2026 by

Example 4 - Show f(x) = {x3+3, 1 is not continuous at x = 0 - Examples

part 2 - Example 4 - Examples - Serial order wise - Chapter 5 Class 12 Continuity and Differentiability
 

 

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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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