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Sketch the graph of f, then identify the values of x0 for which limx→x0 f(x) exists. f(x) = ,,,{x2,x≤28-2x,2<x<44,x≥4

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प्रश्न

Sketch the graph of f, then identify the values of x0 for which `lim_(x -> x_0)` f(x) exists.

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`

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उत्तर

f(x) = `{{:(x^2",", x ≤ 2),(8 - 2x",", 2 < x < 4),(4",", x ≥ 4):}`

x 0 1 2 3 3.5 4 5 6
f(x) x2 x2 x2 8 – 2x 8 – 2x 4 4 4
f(x) 0 1 4 2 1 4 4 4


At x = 4, the curve does not exist.

Hence, except at x0 = 4, the limit of f(x) exists.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Calculus - Limits and Continuity - Exercise 9.1 [पृष्ठ ९७]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 9 Differential Calculus - Limits and Continuity
Exercise 9.1 | Q 16 | पृष्ठ ९७

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