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Evaluate the following : limx→0[ex+e-x-2x⋅tanx]

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Question

Evaluate the following :

`lim_(x -> 0)[("e"^x + "e"^-x - 2)/(x*tanx)]`

Sum
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Solution

`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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Chapter 7: Limits - Miscellaneous Exercise 7.2 [Page 159]

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