I need help finding this limit: lim x → 1 sin ⁡ π x sin...

Yesenia Obrien

Yesenia Obrien

Answered

2022-07-03

I need help finding this limit:
lim x 1 sin π x sin 3 π x

Answer & Explanation

Caiden Barrett

Caiden Barrett

Expert

2022-07-04Added 20 answers

Recall that
sin ( A + B ) = sin A cos B + cos A sin B
We have
lim x 1 sin π x sin 3 π x = lim x 1 sin π x sin ( 2 π x + π x ) = lim x 1 sin π x sin ( 2 π x ) cos ( π x ) + cos ( 2 π x ) sin ( π x ) = lim x 1 sin π x 2 sin ( π x ) cos ( π x ) cos ( π x ) + cos ( 2 π x ) sin ( π x ) = lim x 1 1 2 cos ( π x ) cos ( π x ) + cos ( 2 π x ) = 1 2 cos ( π ) cos ( π ) + cos ( 2 π ) = 1 3
aggierabz2006zw

aggierabz2006zw

Expert

2022-07-05Added 5 answers

lim x 1 sin π x sin 3 π x .
To solve this problem I recommend to use the Taylor series:
(1) sin ( x ) = ( x x 3 3 ! + x 5 5 ! + x 7 7 ! + . . . . ) = n = 0 ( 1 ) n x 2 n + 1 ( 2 n + 1 ) ! .
(2) sin ( α + π ) = sin ( α ) .
we need to perform a substitution x = ( y + 1 ), therefore, y 0
Using the very well known formula (2) and the substitution x = ( y + 1 ) relatively our problem, therefore, we will have:
lim y 0 sin π y sin 3 π y = lim y 0 ( π y ( π y ) 3 3 ! + ( π y ) 5 5 ! + ( π y ) 7 7 ! + . . . . ) ( 3 π y ( 3 π y ) 3 3 ! + ( 3 π y ) 5 5 ! + ( 3 π y ) 7 7 ! + . . . . ) = = lim y 0 y ( π π 3 y 2 3 ! + π 5 y 4 5 ! + π 7 y 6 7 ! + . . . . ) y ( 3 π ( 3 π ) 3 y 2 3 ! + ( 3 π ) 5 y 4 5 ! + ( 3 π ) 7 y 6 7 ! + . . . . ) = lim y 0 π 3 π = 1 3 .
Therefore,
lim x 1 sin π x sin 3 π x = 1 3 .

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