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Examine the following function for continuity: f(x) = ЁЭСе2тИТ25/ЁЭСе+5, x тЙа тИТ5 - Mathematics

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Examine the following function for continuity:

f(x) = `(x^2 - 25)/(x + 5)`, x ≠ −5

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

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

= x − 5

тИ╡ 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

тИ╡ `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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рдкрд╛рда 5: Continuity and Differentiability - Exercise 5.1 [рдкреГрд╖реНрда резрелреп]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдкрд╛рда 5 Continuity and Differentiability
Exercise 5.1 | Q 1.3 | рдкреГрд╖реНрда резрелреп

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