求解当x趋向0时lim[sin(2^x)ln(x+1)]/(2^x-1)

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求解当x趋向0时lim[sin(2^x)ln(x+1)]/(2^x-1)

lim [sin(2^x)ln(x+1)]/(2^x - 1) = lim sin(2^x) * [ lim ln(x+1) / (2^x - 1) ] = sin1 * lim {1/[(x+1)* 2^x * ln2] } = sin1 / ln2

用罗比达法则,结果=0