Wednesday, 18 March 2015

limits - How to find limxto0fraccos(ax)cos(bx)cos(cx)sin(bx)sin(cx)



How to find limx0cos(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((bc)x)cos((b+c)x)cos((bc)x)cos((b+c)x).



Then by repeated application
2asin(ax)(bc)sin((bc)x)(b+c)sin((b+c)x)(bc)sin((bc)x)(b+c)sin((b+c)x),


and
2a2cos(ax)(bc)2cos((bc)x)(b+c)2cos((b+c)x)(bc)2cos((bc)x)(b+c)2cos((b+c)x).



The limit is




2a2(bc)2(b+c)2(bc)2(b+c)2=a2b2c22bc.


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