Use L'Hopital's rule to find limit x(pi/2 -tan^-1x) as x approaches infinity
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Expert's answer
2012-07-26T08:19:49-0400
limit x(pi/2 -tan^-1x)x approaches infinity=limit (pi/2 -tan^-1x)/(1/x)x approaches infinity Both functions 1/x and(pi/2 -tan^-1x) tend to 0 when x tends to infinity/ So we use L'Hopital's rule. (1/x)'=-1/x^2 (pi/2 -tan^-1x)'=-1/(1+x^2) limit (pi/2 -tan^-1x)/(1/x)x approaches infinity=limit {-1/(1+x^2)/(-1/x^2)}x approaches infinity= =limit {x^2/(1+x^2)}x approaches infinity=1 To know more about L'Hopital's rule you can watch our video which contains examples and detailed explanations. Take a look!
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