Ex 13.1, 30 - If f(x) = { |x| + 1, 0, |x| - 1 - Chapter 13 - Ex 13.1

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  1. Chapter 13 Class 11 Limits and Derivatives
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Ex 13.1, 30 If f(x) = x﷯+1, x<0﷮0 x=0﷮ x﷯ −1, x>0﷯﷯ . For what value (s) of a does 𝑙𝑖𝑚﷮𝑥→𝑎﷯ f(x) exists? We need to find value of a for which lim﷮x→a﷯ f(x) exists lim﷮x→a﷯ f(x) exists only if lim﷮ x→a﷮+﷯﷯f (x) = lim﷮ x→a﷮−﷯﷯ f(x) We check limit different values of a • When a = 0 • When a < 0 • When a > 0 Case 1: When a = 0 Limit exists at a = 0 if lim﷮x→ 0﷮+﷯﷯ f(x) = lim﷮x→ 0﷮−﷯﷯ f(x) f(x) = x﷯+1, x<0﷮0 x=0﷮ x﷯ −1, x>0﷯﷯ . Since 1 ≠ – 1 ∴ lim﷮ x→0﷮−﷯﷯f(x) ≠ lim﷮ x→0﷮+﷯﷯f(x) So, left hand limit and right hand limit are not equal at x = 0 Hence lim﷮x→0﷯ f(x) does not exists ∴ At x = 0 Limit does not exists Case 2: When x = a, a > 0 Limit exists at a if lim﷮x→ 𝑎﷮+﷯﷯ f(x) = lim﷮x→ 𝑎﷮−﷯﷯ f(x) f(x) = x﷯+1, x<0﷮0 x=0﷮ x﷯ −1, x>0﷯﷯ . Since lim﷮ x→𝑎﷮−﷯﷯f(x) = lim﷮ x→𝑎﷮+﷯﷯f(x) Hence at all point x = a, a > 0 𝐥𝐢𝐦﷮𝐱→𝒂﷯ f(x) exists Case 3: When x = a, a < 0 Limit exists at a if lim﷮x→ 𝑎﷮+﷯﷯ f(x) = lim﷮x→ 𝑎﷮−﷯﷯ f(x) f(x) = x﷯+1, x<0﷮0 x=0﷮ x﷯ −1, x>0﷯﷯ . Since lim﷮ x→𝑎﷮−﷯﷯f(x) = lim﷮ x→𝑎﷮+﷯﷯f(x) Hence at all point x = a, a < 0 𝐥𝐢𝐦﷮𝐱→𝒂﷯ f(x) exists ∴ 𝐥𝐢𝐦﷮𝐱→𝒂﷯ f(x) exists exists for all a ≠ 0

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