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प्रश्न
Discuss the continuity of the following function at the point(s) or in the interval indicated against them.
`f(x) = (3^x + 3^-x - 2)/x^2` for x ≠ 0.
= (log3)2 for x = 0 at x = 0
उत्तर
f(0) = (log3)2 ...(given)
`lim_(x→0) "f"(x) = lim_(x→0) (3^x + 3^-x - 2)/x^2`
= `lim_(x→0) (3^x + 1/3^x - 2)/x^2`
= `lim_(x→0) ((3^x)^2 + 1 - 2(3^x))/(x^2 . (3)^x)`
= `lim_(x→0) (3^x - 1)^2/(x^2 . (3)^x)` ...[∵ a2 - 2ab + b2 = (a - b)2]
= `lim_(x→0) [((3^x - 1)/x)^2 xx 1/3^x]`
= `lim_(x→0) ((3^x - 1)/x)^2 xx 1/(lim_(x→0)3^x)`
= `(log 3)^2 xx 1/3^0 ....[because lim_{x→0}(("a"^"n" - 1)/x) = log "a"]`
= `(log 3)^2 xx 1/1`
= `(log 3)^2`
∴ `lim_(x→0) "f"(x) = f(0)`
∴ f is continuous at x = 0
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