Question

What is the limit as x approaches 1+ of the function (lnx)^x-1?

It is in the form 0^0, but how do I rewrite it so it is in 0/0 or inf/inf form?

Answer 1

`lim_(x->1^+)(lnx)^(x-1)=1`

Explanation:

Call the limit L:

`L=lim_(x->1^+)(lnx)^(x-1)`

Take the natural logarithm of both sides:

`ln(L)=lim_(x->1^+)ln((lnx)^(x-1))`

Now using the properties of logarithms, we can express the exponent as a product of the `ln`

`ln(L)=lim_(x->1^+)(x-1)*ln(lnx)`

Now we can express this as

`ln(L)=lim_(x->1^+)(ln(lnx))/(1/(x-1))`

which is of the form `(-oo)/oo`

Note that as `x->1^+`, `lnx->0^+`

Therefore `ln(lnx)->-oo`

We can now use L'Hôpital's Rule

`lim_(x->1^+)(ln(lnx))/(1/(x-1))=lim_(x->1^+)(1/(xlnx))/(-1/(x-1)^2)=lim_(x->1^+)(-(x-1)^2)/(xlnx)`

Which is now of the form `0/0`

Let's try L'Hôpital one more time:

`lim_(x->1^+)(-(x-1)^2)/(xlnx)=lim_(x->1^+)(-2(x-1))/(lnx+1)=0/1=0`

Now recall that we took the natural log of the original limit, so

`ln(L)=0`

Therefore,

`L=e^0=1`