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Question
Evaluate the following limits, if necessary use l’Hôpital Rule:
`lim_(x -> x/2) secx/tanx`
Solution
`lim_(x -> x/2) secx/tanx` .......`[oo/oo "Indeterminate form"]`
Simplifying, we get
`lim_(x -> x/2) 1/sinx`
= `1/(sin pi/2)`
= 1
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