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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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