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Calculus problems on finding limits?
Can someone explain how to find the limit of 6tan(2x) / 7x as x -> 0 and 5xcsc(2x) / cos(5x) as x -> 0
1 Answer
- 8 years ago
Using L'Hôpital's rule,
1. lim 6tan(2x)/7x as x -> 0
= lim d/dx[(6tan(2x)]/d/dx[7x] as x -> 0
= lim 12sec²(2x)/7 as x -> 0
= 12sec²(0)/7
= 12/7
2. lim 5xcsc(2x)/cos(5x) as x -> 0
= lim 5x/sin(2x)cos(5x)
= lim d/dx[5x]/d/dx[sin(2x)cos(5x)] as x -> 0
= lim 5 / [-5sin(2x)sin(5x) + 2cos(5x)cos(2x)] as x -> 0
= 5/ [-5sin(0)sin(0) + 2cos(0)cos(0)] as x -> 0
= 5/2
