constant to the power of x???

**runner07** · November 30th 2007, 07:51 PM

so here's the problem....

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

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

**kalagota** · November 30th 2007, 07:58 PM

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..

**runner07** · November 30th 2007, 08:00 PM

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

**kalagota** · November 30th 2007, 08:09 PM

Originally Posted by runner07

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

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

**runner07** · November 30th 2007, 08:22 PM

okay. thanks !

**Krizalid** · December 1st 2007, 04:24 AM

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

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

**galactus** · December 1st 2007, 05:31 AM

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:

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

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

**DivideBy0** · December 1st 2007, 06:12 AM

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

**topsquark** · December 1st 2007, 06:19 AM

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

**ThePerfectHacker** · December 1st 2007, 02:32 PM

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?

**galactus** · December 1st 2007, 02:36 PM

So why do all this fancy stuff?

Because it's fun and we can. Occam's Razor be damned.

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