$\lim \limits_{x \to e} (1-\log x)\log (x-e)$ without l'hopital's rule $\lim \limits_{x \to e}\space\space(1-\log x)\log  (x-e)$
How can is it possible to eliminate  indeterminate without using  L'Hospital's Rule ? I tried to manipulate this formula but still the same problem.

Let put $$x=te$$ and compute the limit
$$\lim_{t\to1^-}(1-log(te))log(e-te)$$
$$=-\lim_{t\to1^-}log(t)(1+log(1-t))$$
$$=-\lim_{t\to1^-}log(1+(t-1))(1+log(1-t))$$
$$=-\lim_{t\to1^-} \frac{log(1+(t-1))}{t-1}(t-1+(t-1)log(1-t))$$
$=-1(0-0)=0$.
