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Examine the following function for continuity: f(x)=x2-25x+5,x≠-5 - Mathematics

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प्रश्न

Examine the following function for continuity:

`f(x) = (x^2 - 25)/(x + 5), x != -5`

योग

उत्तर

Let `f (x) = (x^2 - 25)/(x + 5)`

`= ((x + 5) (x - 5))/(x + 5)`

= x - 5

∵ f (x) = x - 5

Let 'a' be a real number, then,

`lim_(x->a^-) f(x) = lim_(h->0) (a + h) - 5 = a - 5`

`lim_(x->a^+) f(x) = lim_(h->0) (a - h) - 5 = a - 5`

Also, f(a) = a - 5

∵`lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(a)`

Hence, the given function f(x) = x - 5 is continuous at every point of its domain.

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अध्याय 5: Continuity and Differentiability - Exercise 5.1 [पृष्ठ १५९]

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एनसीईआरटी Mathematics [English] Class 12
अध्याय 5 Continuity and Differentiability
Exercise 5.1 | Q 1.3 | पृष्ठ १५९

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