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प्रश्न
Evaluate the following limits:
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx`
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उत्तर
We know `lim_(x -> 0) ("e"^x - 1)/x` = 1
`lim_(x -> 0) sinx/x` = 1
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx = lim_(x -> 0) ("e"^x - 1/"e"^x)/sinx`
= `lim_(x -> 0) (("e"^x * "e"^x - 1)/"e"^x)/(sinx)`
= `lim_(x -> 0) ("e"^(2x) - 1)/("e"^x sinx)`
= `lim_(x -> 0) (1/"e"^x xx ("e"^(2x) - 1)/(1/2 xx 2x) xx x/sinx)`
= `(lim_(x -> 0) 1/"e"^x) 2(lim_(2x -> 0) ("e"^(2x) - 1)/(2x)) xx 1/((lim_(x -> 0) sinx/x))`
= `1/"e"^0 xx 2 xx 1 xx 1/1`
= `1/1 xx 2 xx 1`
`lim_(x -> 0) ("e"^x - "e"^(-x))/sinx` = 2
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