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प्रश्न
Examine the following function for continuity:
`f(x) = (x^2 - 25)/(x + 5), x != -5`
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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