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