so here's the problem....
lim [ (2^x - 1) / x ] as x-> 0
i never get these right. how would you do this?
More generally:
Let $\displaystyle a>0.$
$\displaystyle \lim_{x\to0}\frac{a^x-1}x=\ln a.$
Let's set a change of variables according to $\displaystyle u=\frac1{a^x-1}\implies x=\log_a\left(1+\frac1u\right)\,\therefore\,u\to\i nfty,$ this yields
$\displaystyle
\begin{aligned}
\mathop {\lim }\limits_{x \to 0} \frac{{a^x - 1}}
{x} &= \mathop {\lim }\limits_{u \to \infty } \frac{1}
{{u \cdot \log _a \left( {1 + \dfrac{1}
{u}} \right)}}\\
&= \mathop {\lim }\limits_{u \to \infty } \frac{1}
{{\log _a \left( {1 + \dfrac{1}
{u}} \right)^u }}.
\end{aligned}
$
Which finally yields
$\displaystyle \mathop {\lim }\limits_{u \to \infty } \frac{1}
{{\dfrac{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }}
{{\ln a}}}} = \mathop {\lim }\limits_{u \to \infty } \frac{{\ln a}}
{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }} = \ln a\,\blacksquare$
The Big K always has a nice approach. Here's another.
$\displaystyle \lim_{x\rightarrow{0}}\frac{2^{x}-1}{x}$
Let $\displaystyle u=2^{x}-1, \;\ x=\frac{ln(u+1)}{ln(2)}$
Upon making the subs:
$\displaystyle ln(2)\lim_{u\rightarrow{0}}\frac{u}{ln(u+1)}$
Now, using the first few terms of the Taylor expansion for $\displaystyle ln(u+1)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}.......$ and simplifying, we get:
$\displaystyle ln(2)\lim_{u\rightarrow{0}}\frac{6}{2u^{2}-3u+6}$
$\displaystyle ln(2)(1)=ln(2)$
Wow, how do you guys know to do all these wacky substitutions? Are these tricks you've memorized? Or like, do you think 'Let's make a sub in there and see what happens next', or can you see it all straight away? For instance with Krizalid's sub I would never have seen it coming in a hundred years
Kriz typically comes up with some fantastic ways of looking at this stuff. For the rest of us (not usually me when it comes to limits) it's a function of having done many of them before and a bit of "I've seen something similar to this before and this substitution worked..."
-Dan