Hi everyone, there is a limit problem I am trying to solve for an hour now.
I need to solve it without using l'Hôpital's rule because we haven't covered it in the course yet.
I now that solution is -1, I just can't find the way.
Observe that $\displaystyle \left(\frac{1}{2}\right)^x=2^{-x}$. So we have the following:
$\displaystyle \begin{aligned}\lim\limits_{x\to0}\frac{2^x-1}{2^{-x}-1} &= \lim\limits_{x\to 0}\frac{2^x(2^x-1)}{2^x(2^{-x}-1)}\\ &= \lim\limits_{x\to0}\frac{2^x(2^x-1)}{1-2^x}\\ &= \ldots\end{aligned}$
Something nice will happen when you continue simplifying this limit...
Is $\displaystyle 2^{x}= e^{x\ \ln 2}$ and $\displaystyle 2^{-x}= e^{-x\ \ln 2}$ so that...
$\displaystyle \displaystyle \frac{e^{x\ \ln 2}-1}{e^{-x\ \ln 2}-1}= \frac{e^{\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2}}\ \frac{e^{\frac{x}{2}\ \ln 2} - e^{-\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2} - e^{\frac{x}{2}\ \ln 2}}= - \frac{e^{\frac{x}{2}\ \ln 2}}{e^{-\frac{x}{2}\ \ln 2}}$
... and now the task is easy...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$