10: This expression is of the type
0⋅∞⋅∞
(for details, see previous examples of similar type). This is an
indeterminate product
and no algebraic trick seems possible, so the standard approach should be
used: Change the product into a ratio. However, there are three terms now.
What to do?
There are essentially two possibilities. One is to take the whole infinity
part "nln(n)" and "put it under". The advantage is that the
resulting ratio will be for sure of the type "zero over zero", so the
l'Hospital rule could be used; that is, there is no catch waiting (at least
as far as we can see now). The disadvantage is that the expression that will
be created in the denominator will be far from nice, and taking derivative
will only make it worse.
What is the alternative? Take one "infinity" away, that is, split the limit
into two. For instance, take the logarithm out, the limit of the logarithm is
infinity. The second limit will be the arctangent part multiplied by
n, which is still an indeterminate product, but now much simpler.
Indeed, moving n "under" will create
n−1 there,
which is easy to differentiate. What is the disadvantage of this approach?
After evaluating the two limits, the answers would have to be put together.
The first limit is infinity, so if the second comes out as infinity or some
number not equal to zero, the limit algebra could be used to put the
results together and all is well. However, if the second limit happens to be
zero, the two results would make together an indeterminate product, so all
the work would be in vain and one would have to try something else, for
instance the first procedure outlined above.
Choose whichever way you feel like. The author is a chicken and decided to
play it safe, the hints will follow the less risky (albeit perhaps longer)
first approach. We will briefly outline the shorter but risky way as a
note in the next hint. Try it, try also to pull out the n, it's a good
practice.
Next hint
Answer