Ex 13.1, 26 - Evaluate lim x->0 f(x), f(x) = { x/|x|, 0 - Ex 13.1

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  1. Chapter 13 Class 11 Limits and Derivatives
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Ex 13.1, 26 (Method 1) Evaluate lim﷮x→0﷯ f(x), where f(x) = 𝑥﷮ 𝑥﷯﷯﷮0,﷯﷯, x≠0﷮x=0﷯ Finding limit at x = 0 lim﷮x→ 0﷮−﷯﷯ f(x) = lim﷮x→ 0﷮+﷯﷯ f(x) = lim﷮x→0﷯f(x) Thus, lim﷮ x→0﷮−﷯﷯f(x) = –1 & lim﷮ x→0﷮+﷯﷯f(x) = 1 Since –1 ≠ 1 So, 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ f(x) ≠ 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮+﷯﷯ f(x) So, left hand limit & right hand limit are not equal Hence, 𝒍𝒊𝒎﷮𝐱→𝟎﷯ f(x) does not exist Ex13.1, 26 (Method 2) Evaluate lim﷮x→0﷯ f(x), where f(x) = x﷮ x﷯﷯﷮0,﷯﷯, x≠0﷮x=0﷯ We know that lim﷮x→a﷯ f(x) exist only if lim﷮ x→𝑎﷮−﷯﷯f(x) = lim﷮ x→𝑎﷮+﷯﷯f(x) Similarly in this question we have find limits First we have to prove limit exists by proving lim﷮ x→0﷮+﷯﷯f(x) = lim﷮ x→0﷮−﷯﷯f(x) For 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮+﷯﷯ f(x) f(x) = 𝑥﷮ x﷯﷯ So, as x tends to 0, f(x) tends to 1 ∴ 𝑙𝑖𝑚﷮ 𝑥→0﷮+﷯﷯ f(x) = 1 For 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ f(x) f(x) = 𝑥﷮ x﷯﷯ So, as x tends to 0, f(x) tends to –1 ∴ 𝑙𝑖𝑚﷮ 𝑥→0﷮−﷯﷯ f(x) = –1 Thus, lim﷮ x→0﷮−﷯﷯f(x) = –1 & lim﷮ x→0﷮+﷯﷯f(x) = 1 Since –1 ≠ 1 So, 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮−﷯﷯ f(x) ≠ 𝒍𝒊𝒎﷮ 𝐱→𝟎﷮+﷯﷯ f(x) So, left hand limit & right hand limit are not equal Hence, 𝒍𝒊𝒎﷮𝐱→𝟎﷯ f(x) does not exist

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.