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Question
Using L ‘Hospital’s Rule, evaluate:
`lim_("x"->pi/2) ("x" "tan""x" - pi/4 . "sec" "x")`
Solution
`lim_("x"->pi/2) ["x" "tan""x" - pi/4 . "sec" "x"]`
`= lim_(x->pi/2)[("x sin x" - pi/4)/"cos x"]`
`= lim_(x->pi/2) [("x cos x" + "sin x".1 - 0)/-"sin x"]`
`= lim_(x->pi/2) = ("x cos x" + "sin x")/-"sin x"`
`= (0 + 1)/-1 = -1`
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