Find the limx→∞(lnx−x).
We know that lnx=o(x) as x→∞ therefore we can guess that the limit will be −∞.
Intuitively x goes to infinity way faster than lnx.
Here it is my formal proof of this:
We have that limx→∞x2lnx=+∞ thus form the definition, ∃a>0 such that x>2lnx forall x>a.
Now from this,∀x>a, we deduce that x−lnx>2lnx−lnx=lnx⇒lnx−x<−lnx
Finally we have lim supx→∞(lnx−x)⩽−∞
Thus lim supx→∞(lnx−x)=lim infx→∞(lnx−x)=−∞ .
Is my argument correct?
Thank you in advance!
Answer
for x>0,
ln(x)−x=x(ln(x)x−1)
and limx→+∞ln(x)x=0
thus
lim+∞(ln(x)−x)=−∞
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