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प्रश्न
Evaluate : `lim_(x->0)((2x+1)/(x+1))^(1/x)`
योग
उत्तर
let L`=lim_(x->0)((2x+1)/(x+1))^(1/x)`
Take log on both sides,
`therefore log "L"=lim_(x->0)1/x((2x+1)/(x+1))`
`=lim_(x->0)((2x+1)/(x+1))`
Apply L’Hospital rule ,
`therefore log "L"=lim_(x->0)(2/(2x+1))`
`therefore log "L"=2`
`therefore "L"=e^2`
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L‐ Hospital Rule
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