Use L’Hopitals Rule to evaluate the following limits lim (ln 𝑥)^𝑥−1 𝑥→1

Question

Use L’Hopitals Rule to evaluate the following limits lim _𝑥→1(ln 𝑥)^𝑥−1

Answer 1

Sick Problem:

y = (lnx)^x-1

Let's do the limit from Y = ln(y) and use ln(xⁿ) = nln(x)

Lim (x->1) (x-1)ln(lnx)) which we write as ln(ln(x))/ (1/(x-1)) so that we have a -inf/inf situation and can apply l'Hopital's

So Taking the derivative of num and denom wrt x:

1/(ln(x)) * 1/x or 1/(xln(x)) and denom is -1/(x-1)² which leaves the fraction as

lim as x ->1 of -(x-1)²/(xln(x)) (Still 0/0, L'Hopital's again)

lim as x ->1 of 2(x-1)/ (ln(x)+1) which approaches 0 by plug-in)

If Y--> 0, then y --> 1