This morning a colleague asked me how I would solve quickly this limit, lim x...

pouzdrotf

pouzdrotf

Answered

2022-07-03

This morning a colleague asked me how I would solve quickly this limit,
lim x π 4 cos ( 2 x ) 2 cos ( x ) 2

Answer & Explanation

Sophia Mcdowell

Sophia Mcdowell

Expert

2022-07-04Added 14 answers

Consider rewriting the numerator as cos ( 2 x ) cos ( π 2 ) . Then we can use the same manipulation to get
cos ( 2 x ) cos ( π 2 ) = 2 sin ( 2 x + π 2 2 ) sin ( 2 x π 2 2 ) = 2 sin ( x + π 4 ) sin ( x π 4 )
Substituting this in gives us
lim x π 4 2 sin ( x + π 4 ) sin ( x π 4 ) 2 ( 2 sin ( x + π 4 2 ) sin ( x π 4 2 ) ) = 1 2 lim x π 4 sin ( x + π 4 ) sin ( x + π 4 2 ) lim x π 4 sin ( x π 4 ) sin ( x π 4 2 )
supposing for the moment that those two limits exist. (also note the additional factor of 2 in the denominator which seems to have been dropped in your manipulation)
For the first limit we can simply evaluate at x = π 4 to get 1 1 2 = 2 .
For the second limit, consider rewriting sin ( x π 4 ) in the numerator as sin ( 2 x π 4 2 ) = 2 sin ( x π 4 2 ) cos ( x π 4 2 ) . This gives us that
lim x π 4 sin ( x π 4 ) sin ( x π 4 2 ) = lim x π 4 2 cos ( x π 4 2 ) = 2
So our total limit is 1 2 2 2 = 2 , which agrees with your initial solution.
Wade Bullock

Wade Bullock

Expert

2022-07-05Added 5 answers

If you set x = u + π 4 with u 0 then you get
f ( x ) = cos ( 2 x ) 2 cos ( x ) 2 = 2 sin ( u ) cos ( u ) 1 2 ( cos ( u ) 1   sin ( u ) sin ( u ) ) 2 sin ( u ) 2 sin ( u ) 2
The fact that cos ( u ) 1 sin ( u ) can be inferred from:
| cos ( u ) 1 sin ( u ) | = | 2 sin ( u / 2 ) 2 sin ( u ) | = | sin ( u / 2 ) 2 u 2 / 4 | 1 × | u sin ( u ) | 1 × | 8 u | 0 0

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