Question

`lim_(x->1)(x^x-1)/(log_ex-x+1)`?

Answer 1

The limit doesn't exist.

Explanation:

The limit is indeterminate in the form `0//0` so L'Hôpital's rule applies. Take the numerator of the numerator and denominator separately.

To take the derivative of `x^x`, note that `x^x=e^(ln(x^x))=e^(xlnx)`. So, `d/dxx^x=d/dxe^(xlnx)=e^(xlnx)(1*lnx+x*1/x)=x^x(lnx+1)`.

(I'm using the notation `log_ex=lnx`.)

Then,

`lim_(xrarr1)(x^x-1)/(lnx-x+1)=lim_(xrarr1)(x^x(lnx+1))/(1/x-1)`

As this is in the form `1//0`, we can say that the limit doesn't exist.

A graph of the function reveals that the function is asymptotic as `x` approaches `x=1`.