$\displaystyle \forall a>0$, prove that $\displaystyle \lim_{x\to0}\frac{a^x-1}x=\ln a$
I like this limit, 'cause most of people the first thing they do is apply L'Hôpital's Rule.
$\displaystyle \forall a>0$, prove that $\displaystyle \lim_{x\to0}\frac{a^x-1}x=\ln a$
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.
$\displaystyle a^x = e^{x\ln a} = 1+(x\ln a) + \frac{(x\ln a)^2}{2!}+...$
Thus,
$\displaystyle \frac{a^x-1}{x} = \ln a + \frac{x\ln^2 a}{2!}+\frac{x^2\ln a}{3!}+...$.
Since power series are uniformly convergent we can pass the limit through to get,
$\displaystyle \ln a + 0 + 0 +... = \ln a$.
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Solution #2.
This is just the derivative of $\displaystyle a^x$ at $\displaystyle 0$.
$\displaystyle a^x-1=u\Rightarrow x=\frac{\ln(1+u)}{\ln a}$
Then $\displaystyle \displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=\lim_{u\to 0}\frac{u\ln a}{\ln(1+u)}=$
$\displaystyle \displaystyle =\ln a\lim_{u\to 0}\frac{1}{\ln(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\displaystyle\ln\lim_{u\to 0}(1+u)^{\frac{1}{u}}}=\frac{\ln a}{\ln e}=\ln a$
Okay
Here's another solution similar to red_dog's one
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
$\displaystyle \mathop {\lim }\limits_{u \to \infty } \frac{1}
{{\dfrac{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }}
{{\ln a}}}} = \mathop {\lim }\limits_{u \to \infty } \frac{{\ln a}}
{{\ln \left( {1 + \dfrac{1}
{u}} \right)^u }} = \ln a$