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Find the limit: lim as x->0+: x^(5.7)lnx
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A nice comparison seems reasonable. \(x > ln(x)\;and\;x^{5}>x^{5.7}\;for\;x < 1\) Have you seen L'Hospital?
@tkhunny That's what I am looking at right now
Well, then, we should be able to find the solution that way, too. Honestly, thought, this one is fine with the comparison. We can sandwich the thing with little difficulty. First, we need to know the L'Hospital's Rule is applicable. Do you recall the criteria?
@tkhunny I got it!
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How did you determine the applicability of the rule?
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