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प्रश्न
Evaluate the following limits, if necessary use l’Hôpital Rule:
`lim_(x -> 0) (1/sinx - 1/x)`
उत्तर
`lim_(x -> 0) (1/sinx - 1/x)` .....`[oo - oo "Indeterminate form"]`
= `lim_(x -< 0) (x - sin x)/(xsinx)` .......`[0/0 "Indeterminate form"]`
Applying L' Hôpital's rule,
= `lim_(x -> 0) (1 - cosx)/(xcosx + sin x.1)` .......`[0/0 "Indeterminate form"]`
Again Applying L' Hôpital's rule,
= `lim_(x -> 0) sinx/(-xsinx + cosx + cosx)`
= `lim_(x -> 0) sinx/(-xsinx + 2cosx)`
= `0/2`
= 0
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