1. ## [SOLVED] Limit

Hello
I'm having difficulties with this limit:

$\displaystyle \lim_{x\rightarrow e^{-}}(ln x)^{\frac{1}{ln^{2}ln x}}$

Thanks for help

2. ## Re: Limit

\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*}

3. ## Re: Limit

That's the same pattern that I followed, but on my book the solution to that limit is 0 ...I'm starting to think that's a typo

p.s. thanks for the fast reply

4. ## Re: Limit

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*}.

5. ## Re: Limit

That's right, got it Have a good day