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

Question

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

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Answer 1 · 2018-03-28T19:44:41

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

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,

$L=e^0=1$