Question

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` ?

Answer 1

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`

Answer 2

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`