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