L = x → 0 lim e 2 x − 1 e x a x − ( a x + e x ) + ln ( 1 + x ) e x − 1 − e x sin x + sin x ( 1 + 2 e x sin x ) + e x cos ( 2 x )
Let a be a constant positive real number , find the value of L .
Clarification : e denotes Euler's number , e ≈ 2 . 7 1 8 2 8 .
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Let function in numerator be f ( x ) and in denominator be g ( x ) .
By using L'Hospital rule :
Now L i m i t r e q = L t x − > 0 f ( x ) / g ( x ) = L t x − > 0 f ′ ( x ) / g ′ ( x ) .
Since this argumant of limit is not of form 0/0 , we find
L r e q = f ′ ( 0 ) / g ′ ( 0 ) = 0 . 5 .
In the third line it should be (1-2(sinx)^2) Probably you forgot to remove e^x :)
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L = x → 0 lim e 2 x − 1 e x a x − ( a x + e x ) + ln ( 1 + x ) e x − 1 − e x sin x + sin x ( 1 + 2 e x sin x ) + e x cos ( 2 x ) = x → 0 lim e 2 x − 1 e x a x − a x − e x + ln ( 1 + x ) e x − 1 − e x sin x + sin x + 2 e x sin 2 x + e x cos ( 2 x ) = x → 0 lim ( e x − 1 ) ( e x + 1 ) ( e x − 1 ) a x + ( e x − 1 ) ln ( 1 + x ) − ( e x − 1 ) sin x − e x ( 1 − 2 sin 2 x ) + e x cos ( 2 x ) = x → 0 lim ( e x − 1 ) ( e x + 1 ) ( e x − 1 ) a x + ( e x − 1 ) ln ( 1 + x ) − ( e x − 1 ) sin x − e x cos ( 2 x ) + e x cos ( 2 x ) = x → 0 lim e x + 1 a x + ln ( 1 + x ) − sin x = 1 + 1 1 + 0 − 0 = 2 1