EXAMPLE 6 Evaluate lim x In(x5). x ? 0+ SOLUTION The given limit is indeterminate because, as x? 0+, the first factor (x) approaches while the second factor In(x5) approaches -?o. Writing x = 1/(1/x), we have 1/x ? ? as x ? 0+, so l'Hospital's Rule gives lim_x In(x5) = lim In(x5) x ? 0+ X? 0+ 1/x 5/x = = = lim x ? 0¹-1/x² lim X? 0+

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EXAMPLE 6 Evaluate

lim_{x ? 0^{+}} x ln (x^{5})

. SOLUTION The given limit is indeterminate because, as

x ? 0^{+}

, the first factor

(x)

approaches while the second factor

ln (x^{5})

approaches

? ?

. Writing

x = 1/ (1/ x)

, we have

1/ x ? ?

as

x ? 0^{+}

, so l'Hospital's Rule gives

x ? 0^{+} lim x ln (x^{5}) = x ? 0^{+} lim \frac{ln ( x ^{5} )}{1/ x} = x ? 0^{+} lim \frac{5/ x}{? 1/ x ^{2}}

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