Scary Limit
If the value of the above limit is in the form , where is a integer, then find .
The answer is 2017.
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1 solution
L = x → 0 lim [ sin 2 ( 2 − a x π ) ] sec 2 ( 2 − b x π )
ln ( L ) = x → 0 lim cos 2 ( 2 − b x π ) ln [ sin 2 ( 2 − a x π ) ]
L'hopital applies:
ln ( L ) = x → 0 lim − sin ( 2 − b x 2 π ) ⋅ ( 2 − b x ) 2 b π sin 2 ( 2 − a x π ) 1 ⋅ sin ( 2 − a x 2 π ) ⋅ ( 2 − a x ) 2 a π
ln ( L ) = − b a ⋅ x → 0 lim sin ( 2 − b x 2 π ) sin ( 2 − a x 2 π )
Again, L'hopital applies:
ln ( L ) = − b a ⋅ x → 0 lim cos ( 2 − b x 2 π ) ⋅ ( 2 − b x ) 2 2 b π cos ( 2 − a x 2 π ) ⋅ ( 2 − a x ) 2 2 a π
ln ( L ) = − b 2 a 2
L = e − b 2 a 2
Thus:
k = − 1 , − 2 0 1 7 k = 2 0 1 7