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प्रश्न
If the function f is continuous at = 2, then find f(2) where f(x) = `(x^5 - 32)/(x - 2)`, for ≠ 2.
उत्तर
Consider,
`lim_(x ->2) f(x) = lim_(x->2) [(x^5 - 32)/(x - 2)]`
`lim_(x->2) [(x^5 - 2^5)/(x - 2)]`
= 5 (2)5-1
`(lim_(x->a) (x^n - a^n)/(x -a) = na^n-1)`
= 80
Since f is continuous at x = 2
`lim_(x->2)` f(x) = f(2)
f(2) = 80
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