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प्रश्न
Evaluate the following limit :
`lim_(x -> 0) [(3^x + 3^-x - 2)/(x*tanx)]`
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उत्तर
`lim_(x -> 0) [(3^x + 3^-x - 2)/(xtanx)]`
= `lim_(x -> 0) (3^x[3^x + 3^-x - 2])/(3^x*xtanx)`
= `lim_(x -> 0) ((3^x)^2 + 1 - 2(3^x))/(3^x*xtanx)`
= `lim_(x -> 0) (3^x - 1)^2/(3^x *xtanx)`
= `lim_(x -> 0) ((3^x - 1)/x)^2/(3^x * (tanx/x))` ...[∵ x → 0, x ≠ 0 ∴ x2 ≠ 0]
= `(lim_(x -> 0) (3^x - 1)/x)^2/((lim_(x -> 0) 3^x) xx (lim_(x -> 0) tanx/x)`
= `(log3)^2/(3^0*1) ...[because lim_(x -> 0) ("a"^x - 1)/x = log"a"]`
= (log 3)2
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