How to find limx→0cos(ax)−cos(bx)cos(cx)sin(bx)sin(cx)
I tried using L Hospital's rule but its not working!Help please!
Answer
Before using L'Hospital, turn the products to sums
cos(ax)−cos(bx)cos(cx)sin(bx)sin(cx)=2cos(ax)−cos((b−c)x)−cos((b+c)x)cos((b−c)x)−cos((b+c)x).
Then by repeated application
2asin(ax)−(b−c)sin((b−c)x)−(b+c)sin((b+c)x)(b−c)sin((b−c)x)−(b+c)sin((b+c)x),
and
2a2cos(ax)−(b−c)2cos((b−c)x)−(b+c)2cos((b+c)x)(b−c)2cos((b−c)x)−(b+c)2cos((b+c)x).
The limit is
2a2−(b−c)2−(b+c)2(b−c)2−(b+c)2=−a2−b2−c22bc.
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