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# ?
I computed it the following way:
The above expression
Applying LH Rule to the indeterminate form, we take derivatives of the numerator and denominator and get:
My point is, am I correct in taking the first derivative of
2 Answers
See explanation
Explanation:
Try to reformulate the problem
Note that
which makes makes it difficult to talk about a derivative
A different approach to the problem,
is to consider the limit of
#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
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#
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
(Similarly
In fact, the expression
It is true that for some functions
But we can find
For example,
But