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प्रश्न
Evaluate the following: `lim_(x -> 0) [(3^x + 3^-x - 2)/x^2]`
उत्तर
`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))/(3^x*x^2)`
= `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)`
= `(log3)^2 xx 1/3^0`
= `(log3)^2 xx 1/1 ....[lim_(x -> 0) ("a"^x - 1)/x = log "a"]`
= (log3)2
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