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प्रश्न
Evaluate the following :
`lim_(x -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`
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उत्तर
`lim_(x -> 0) ("e"^x + "e"^-x - 2)/(xtanx)`
= `lim_(x -> 0) ("e"^x+1/"e"^x-2)/(xtanx)`
= `lim_(x -> 0) (("e"^x)^2 + 1 - 2("e"^x))/("e"^x*xtanx)`
= `lim_(x -> 0) ("e"^x - 1)^2/("e"^x *xtanx)`
= `lim_(x -> 0) [(("e"^x - 1)^2/x^2)]/[(("e"^x*xtanx)/x^2)]` ...[∵ x → 0; ∴ x ≠ 0]
= `lim_(x -> 0) (("e"^x - 1)/x)^2/("e"^x*(tanx/x))`
= `(lim_(x -> 0)("e"^x - 1)/x)^2/(lim_(x -> 0)"e"^x*lim_(x -> 0)(tanx/x))`
= `(1)^2/("e"^0*1)`
= 1
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