Ex 5.1, 8 - Find points of discontinuity f(x) = {|x|/x, if x=0

Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.1 ,8 - Chapter 5 Class 12 Continuity and Differentiability - Part 4

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Ex 5.1, 8 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={█(|𝑥|/𝑥, 𝑖𝑓 𝑥≠0@&0 , 𝑖𝑓 𝑥=0)┤ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x > 0 When x < 0 Case 1 : When x = 0 f(x) is continuous at 𝑥 =0 if L.H.L = R.H.L = 𝑓(0) Since there are two different functions on the left & right of 0, we take LHL & RHL . if lim┬(x→0^− ) 𝑓(𝑥)=lim┬(x→0^+ ) " " 𝑓(𝑥)= 𝑓(0) Since L.H.L ≠ R.H.L f(x) is not continuous at x=0 LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) (|−ℎ|)/(−ℎ) = lim┬(h→0) ℎ/(−ℎ) = lim┬(h→0) −1 = −1 RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) (|ℎ|)/ℎ = lim┬(h→0) ℎ/ℎ = lim┬(h→0) 1 = 1 Case 2 : When x < 0 For x < 0, f(x) = (|𝑥|)/𝑥 f(x) = ((−𝑥))/𝑥 f(x) = −1 Since this constant It is continuous ∴ f(x) is continuous for x < 0 (As x < 0, x is negative) Case 3 : When x > 0 For x > 0, f(x) = (|𝑥|)/𝑥 f(x) = 𝑥/𝑥 f(x) = 1 Since this constant It is continuous ∴ f(x) is continuous for x > 0 Hence, only x = 0is point is discontinuity. ∴ f is continuous for all real numbers except 0. Thus, f is continuous for 𝒙 ∈ R − {0} (As x > 0, x is positive)

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