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प्रश्न
Examine the following function for continuity:
`f (x)1/(x - 5), x != 5`
उत्तर
Let a be a real number, then,
`lim_(x->a^+)f(x) = lim_(h->0) 1/(a + h - 5) = 1/(a - 5)`
`lim_(x->a^-) f(x) = lim_(h->0) 1/(a - h - 5) = 1/(a-5)`
`f (a) = 1/(a-5)`
`∵ lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(a)`
Hence, the given function `f (x) = 1/(x - 5)` is continuous at all points except at x = 5.
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