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Using LHopital's Rule, I am to calculate;
(a) (ln(x))/x with limitation of x to infinity
(b) x*ln(x) with limitation of x to 0+
for (a) I did the normal differentiation which gives a 1/x if I'm not wrong. 1/x is a horrible answer to deal with coz in this case, it becomes 1/infinity. Anyone who can clarify?
for (b) I obtained (1+ln(x)) with limit of x to positive. replacing x by positive, I get an incredibly large answer. :Sick
The Limit of ln(x) is indeed 0 for x -> infinity. However, the question asks for the limit of ln(x)/x.
[d/dx][ln(x)/x]
= {x[1/x] - ln(x)}/x^2
=[1 - ln(x)]/x^2
As x -> infinity, ln(x) approaches infinity as well, meaning that the numerator approaches negative infinity. However, as the denominator is x^2, it approaches infinity much faster than the numerator, meaning that the limit is 0- for this function.
As for b)
[d/dx][x*ln(x)]
= x[1/x] + ln(x)
= 1 + ln(x)
As x -> 0+, ln(x) -> negative infinity, hence the limit of b) is negative infinity.
oh my god. what a careless mistake from me in the differentiation.
thx man.
No problem :)