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# Ex 13.1, 24

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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Transcript

Ex 13.1, 24 (Method 1) Find limx→1 f(x), where f(x) = x2 −1,−x2 −1, x ≤1x>1 Finding limit at x = 1 lim x→1− f(x) = lim x→1+ f(x) = limx→1 f(x) Thus, lim x→1+f(x) = –2 & lim x→1−f(x) = 0 ⇒ 𝒍𝒊𝒎 𝐱→𝟏+f(x) ≠ 𝒍𝒊𝒎 𝐱→𝟏−f(x) So, left hand limit & right hand limit are not equal Hence 𝐥𝒊𝒎𝐱→𝟏 f(x) does not exit Ex13.1, 24 (Method 2) Find limx→1 f(x), where f(x) = x2 −1,−x2 −1, x ≤1x>1 We know that limx→𝑎 f(x) exists only if Left Hand limit = Right hand limit i.e. lim x→𝑎+f(x) = lim x→𝑎− f(x) Similarly in question , we have to find limits First we have to prove limx→ 1+ f(x) = limx→ 1− f(x) For 𝒍𝒊𝒎 𝒙→𝟏+f(x) , f(x) = – x2 – 1 Hence if we move value of x towards 1, value of f(x) tends towards – 2 Hence, limx→ 1+ f(x) = – 2 For 𝒍𝒊𝒎 𝒙→𝟏−f(x) , f(x) = x2 – 1 Hence if we move value of x towards 1, value tends of f(x) towards 0 Hence f(x)x→1−= 0 Since – 2 ≠ 0 ∴ 𝒍𝒊𝒎𝐱→ 𝟏+ f(x) ≠ 𝒍𝒊𝒎𝐱→ 𝟏− f(x) So, left hand limit & right hand limit are not equal Hence, limx→1 f(x) dues not exist

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Ex 13.1, 24 You are here

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About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .