Limit without usage of derivatives So I have this limit:
$$\lim\limits_{x\to{-2}}(2\sqrt{-1-x}-1)^{\frac{1}{4^{-x}-16}}$$
So I was thinking on using logarithm and then transform it to use known basic limits. But I am not quite sure how to use logarithm here.
And also I cannot use l'hospital rule or derivatives or even Taylor series. Any help would be appreciated.
 A: Hint: what you need is to use
$$ \lim_{t\to0}(1+t)^{1/t}=e.$$
Here is some more detail:
\begin{eqnarray}
&&\lim\limits_{x\to{-2}}(2\sqrt{-1-x}-1)^{\frac{1}{4^{-x}-16}}\\
&=&\lim\limits_{x\to{-2}}[1-(2-2\sqrt{-1-x})]^{\frac{1}{4^{-x}-16}}\\
&=&\lim\limits_{x\to{-2}}\left\{[1+(2\sqrt{-1-x}-2)]^{\frac{1}{2\sqrt{-1-x}-2}}\right\}^{\frac{2\sqrt{-1-x}-2}{4^{-x}-16}}
\end{eqnarray}
and I think you can do the rest.
Update: letting $t=-(x+2)$
\begin{eqnarray}
\lim_{x\to-2}\frac{2\sqrt{-1-x}-2}{4^{-x}-16}&=&\lim_{t\to0}\frac{\sqrt{1+t}-1}{8(4^{t}-1)}\\
&=&\lim_{t\to0}\frac{\sqrt{1+t}-1}{t}\frac{t}{8(4^{t}-1)}\\
&=&\frac{1}{16\ln4}.
\end{eqnarray}
A: It looks nice to deal with positive numbers so instead of letting $x \to -2$ I put $x = -t$ so that $t \to 2$ and we need to calculate the limit of $$f(x) = (2\sqrt{-1 - x} - 1)^{1/(4^{-x} - 16)} = (2\sqrt{t - 1} - 1)^{1/(4^{t} - 16)} = g(t)$$ as $t \to 2$. Let $L$ be the desired limit then
\begin{align}
\log L &= \log\lim_{t \to 2}g(t)\notag\\
&= \lim_{t \to 2}\log g(t)\text{ (by continuity of log)}\notag\\
&= \lim_{t \to 2}\frac{1}{4^{t} - 16}\log(2\sqrt{t - 1} - 1)\notag\\
&= \lim_{v \to 0}\frac{1}{4^{v + 2} - 16}\log(2\sqrt{v + 1} - 1)\text{ (putting }v = t - 2)\notag\\
&= \frac{1}{16}\lim_{v \to 0}\frac{v}{4^{v} - 1}\cdot\frac{\log(2\sqrt{v + 1} - 1)}{v}\notag\\
&= \frac{1}{16}\cdot\frac{1}{\log 4}\lim_{v \to 0}\frac{\log(1 + 2\sqrt{v + 1} - 2)}{2\sqrt{v + 1} - 2}\cdot\frac{2\sqrt{v + 1} - 2}{v}\notag\\
&= \frac{1}{16\log 4}\lim_{z \to 0}\frac{\log(1 + z)}{z}\cdot\lim_{v \to 0}\frac{2\sqrt{v + 1} - 2}{v}\text{ (putting } z = 2\sqrt{v + 1} - 2)\notag\\
&= \frac{1}{16\log 4}\lim_{v \to 0}\frac{4(v + 1) - 4}{v(2\sqrt{v + 1} + 2)}\notag\\
&= \frac{1}{16\log 4}\lim_{v \to 0}\frac{2}{\sqrt{v + 1} + 1}\notag\\
&= \frac{1}{16\log 4}\notag
\end{align}
Hence $L = \exp\left(\dfrac{1}{16\log 4}\right)$. Thus your "thinking on using logarithm and then transform it to use known basic limits" is the right approach. Anything more than that for this simple problem is an overkill.
