Example 17 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
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Transcript
Example 17 Discuss the continuity of sine function.Let π(π₯)=sinβ‘π₯ Letβs check continuity of f(x) at any real number Let c be any real number. We know that A function is continuous at π₯ = π if L.H.L = R.H.L = π(π) i.e. limβ¬(xβπ^β ) π(π₯)= limβ¬(xβπ^+ ) " " π(π₯)= π(π) LHL at x β c limβ¬(xβπ^β ) f(x) = limβ¬(hβ0) f(c β h) = (πππ)β¬(ββ0) sinβ‘γ(πγββ) = (πππ)β¬(ββ0) (sinβ‘π cosβ‘β "β cos c sin h " ) = (sinβ‘π cosβ‘0 "β cos c sin 0" ) = sinβ‘πΓ 1"β cos c" Γ 0 = sin c sinβ‘(π₯βπ¦) =sinβ‘π₯ cosβ‘π¦βcosβ‘π₯ sinβ‘π¦ π΄π , cosβ‘0=1 & sinβ‘0=0 RHL at x β c limβ¬(xβπ^+ ) f(x) = limβ¬(hβ0) f(c + h) = (πππ)β¬(ββ0) sinβ‘γ(πγ+β) = (πππ)β¬(ββ0) (sinβ‘π cosβ‘β "+ cos c sin h " ) = (sinβ‘π cosβ‘0 "+ cos c sin 0" ) = sinβ‘πΓ 1" + cos c" Γ 0 = sin c sinβ‘(π₯+π¦) =sinβ‘π₯ cosβ‘π¦+cosβ‘π₯ sinβ‘π¦ π΄π , cosβ‘0=1 & sinβ‘0=0 And, π(π) = πππβ‘π Since L.H.L = R.H.L = π(π) Therefore, π(π₯) is continuous for all real number So, πππβ‘π is continuous.
