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प्रश्न
Using L’Hospital Rule, evaluate: `lim_(x->0) (8^x - 4^x)/(4x
)`
उत्तर
`lim_(x-.0) (8^x-4^x)`
`lim_(x->0)(8^xlog8 - 4^x log 4)`
`= (8^0 log 8 - 4^0 log 4)/4`
`=1/4log2`
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