Can we apply L'Hopital's Rule while computing the following limit?

#lim_(nrarr oo) (log0)/n#

I computed it the following way:

The above expression #= (oo)/(oo)# as #n rarr oo#

Applying LH Rule to the indeterminate form, we take derivatives of the numerator and denominator and get:

#lim_(nrarr oo) (log0)/n# = #lim_(nrarr oo)# #0/1# = #0#

My point is, am I correct in taking the first derivative of #log0# to be #0# ?

2 Answers
Feb 24, 2018

See explanation

Explanation:

Try to reformulate the problem

Note that #ln(0)# is undefined,
which makes makes it difficult to talk about a derivative

A different approach to the problem,
is to consider the limit of #ln(x)# when x approaches 0

#lim_(x->0)ln(x)=-oo#

Or alternative

#lim_(x->oo)ln(1/x)=-oo#

So we may reformulate your problem as

#lim_(n->oo)ln(1/n)/n=lim_(n->oo)-ln(n)/n#

Now have the indeterminate form #oo/oo#,
and can apply L'Hopital's Rule

#lim_(n->oo)-ln(n)/n=lim_(n->oo)-(1/n)/1=lim_(n->oo)-1/n=0#

Feb 25, 2018

The answer to your point is "No. You are not correct in thinking that an undefined assembly of symbols (like log 0) has a derivative.

Explanation:

Because #log 0# is not defined, it has no derivative.

(Similarly #7/0# has no derivative. Undefined expressions are not constants.)

In fact, the expression #lim_(nrarroo) log0/n# cannot be defined, because #log0/n# cannot be defined. Since it is undefin ed, there is no sense in which we can "compute" it.

It is true that for some functions #f# with #lim_(nrarroo)f(n) = 0#, we get #lim_(nrarroo) (logf(n))/n = 0#

But we can find #f(n)# such that #lim_(xrarroo)f(n) = 0#, but #lim_(nrarroo)(log(f(n)))/n != 0#

For example,

#lim_(nrarroo) e^(-2n) = 0#

But #lim_(nrarroo) log(e^(-2n))/n = lim_(nrarroo)(-2n)/n = -2#