Hello
I'm having difficulties with this limit:
$\displaystyle \lim_{x\rightarrow e^{-}}(ln x)^{\frac{1}{ln^{2}ln x}}$
Thanks for help
$\displaystyle \displaystyle \begin{align*} \lim_{x \to e^{-}}\left(\ln{x}\right)^{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}} &= \lim_{x \to e^{-}}e^{\ln{\left\{ \left(\ln{x}\right)^{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}} \right\}}} \\ &= \lim_{x \to e^{-}} e^{\left\{\frac{1}{\left[\ln{\left(\ln{x}\right)}\right]^2}\right\}\left[\ln{\left(\ln{x}\right)}\right]} \\ &= \lim_{x \to e^{-}} e^{\frac{1}{\ln{\left(\ln{x}\right)}}} \\ &= \infty \end{align*}$
I know what my mistake was.
$\displaystyle \displaystyle \begin{align*} \frac{1}{X} \to -\infty \end{align*}$ as $\displaystyle \displaystyle \begin{align*} X \to 0 \end{align*}$ from the left.
So $\displaystyle \displaystyle \begin{align*} e^X \to 0 \end{align*}$ as $\displaystyle \displaystyle \begin{align*} X \to -\infty \end{align*}$.