Limit without usage of derivatives

Question

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.

Answer 1

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}

Answer 2

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.