1. ## Known one (limit)

$\forall a>0$, prove that $\lim_{x\to0}\frac{a^x-1}x=\ln a$

I like this limit, 'cause most of people the first thing they do is apply L'Hôpital's Rule.

2. Originally Posted by Krizalid
$\forall a>0$, prove that $\lim_{x\to0}\frac{a^x-1}x=\ln a$

I like this limit, 'cause most of people the first thing they do is apply L'Hôpital's Rule.
I (almost) never used L'Hopital's Rule.

$a^x = e^{x\ln a} = 1+(x\ln a) + \frac{(x\ln a)^2}{2!}+...$
Thus,
$\frac{a^x-1}{x} = \ln a + \frac{x\ln^2 a}{2!}+\frac{x^2\ln a}{3!}+...$.
Since power series are uniformly convergent we can pass the limit through to get,
$\ln a + 0 + 0 +... = \ln a$.

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Solution #2.

This is just the derivative of $a^x$ at $0$.

3. $a^x-1=u\Rightarrow x=\frac{\ln(1+u)}{\ln a}$
Then $\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{u\to 0}\frac{u\ln a}{\ln(1+u)}=$
$\displaystyle =\ln a\lim_{u\to 0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\displaystyle\ln\lim_{u\to 0}(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\ln e}=\ln a$

4. Originally Posted by red_dog
$a^x-1=u\Rightarrow x=\frac{\ln(1+u)}{\ln a}$
Then $\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{u\to 0}\frac{u\ln a}{\ln(1+u)}=$
$\displaystyle =\ln a\lim_{u\to 0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\displaystyle\ln\lim_{u\to 0}(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\ln e}=\ln a$
nice solution!

i believe this is the first time i've seen someone bring a limit inside a log. i didn't know you could do that

5. Originally Posted by Jhevon
i believe this is the first time i've seen someone bring a limit inside a log. i didn't know you could do that
The reason you can do it is that the log function is continuous. That means that if $x\to a$ then $\ln x\to \ln a$.

6. Okay

Here's another solution similar to red_dog's one

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

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