# Thread: constant to the power of x???

1. ## constant to the power of x???

so here's the problem....

lim [ (2^x - 1) / x ] as x-> 0

i never get these right. how would you do this?

2. Originally Posted by runner07
so here's the problem....

lim [ (2^x - 1) / x ] as x-> 0

i never get these right. how would you do this?
are you allowed to use L'hospital's Rule?
use it..

3. yes i know. but how do you derive 2^x. thats my main problem.

4. Originally Posted by runner07
yes i know. but how do you derive 2^x. thats my main problem.

ah, ok... $D_x [a^x] = a^x \ln a$..

5. okay. thanks !

6. Originally Posted by runner07
lim [ (2^x - 1) / x ] as x-> 0
More generally:

Let $a>0.$

$\lim_{x\to0}\frac{a^x-1}x=\ln a.$

Let's set a change of variables according to $u=\frac1{a^x-1}\implies x=\log_a\left(1+\frac1u\right)\,\therefore\,u\to\i nfty,$ this yields


\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

$\mathop {\lim }\limits_{u \to \infty } \frac{1}
{{\dfrac{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }}
{{\ln a}}}} = \mathop {\lim }\limits_{u \to \infty } \frac{{\ln a}}
{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }} = \ln a\,\blacksquare$

7. The Big K always has a nice approach. Here's another.

$\lim_{x\rightarrow{0}}\frac{2^{x}-1}{x}$

Let $u=2^{x}-1, \;\ x=\frac{ln(u+1)}{ln(2)}$

Upon making the subs:

$ln(2)\lim_{u\rightarrow{0}}\frac{u}{ln(u+1)}$

Now, using the first few terms of the Taylor expansion for $ln(u+1)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}.......$ and simplifying, we get:

$ln(2)\lim_{u\rightarrow{0}}\frac{6}{2u^{2}-3u+6}$

$ln(2)(1)=ln(2)$

8. 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

9. Originally Posted by DivideBy0
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

10. Originally Posted by DivideBy0
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
It comes from experience

Note: You can just realize that $\lim (a^x - 1)/x$ is the derivative of $a^x$ at zero. So why do all this fancy stuff?

11. So why do all this fancy stuff?
Because it's fun and we can. Occam's Razor be damned.