Let $a \in \mathbb R$. Use $\lim_{t\rightarrow 0} \frac {\log(1+at)} t = a $ to show $\lim_{t\rightarrow \infty} (1+\frac a t)^t = e^a$ Let $a \in \mathbb R$.
Use the definition of the differential quotient to show: 
$$\lim_{t\rightarrow 0} \frac {\log(1+at)} t = a$$
In my textbook the differential quotient is defined as $$\frac {f(x_0 + h)-f(x_0)} {h}$$
but $f(x_0+h)$ doesn't correspond to the expression $\log(1+at)$ with $x_0 = 1$ ?
How can I utilize the definition for something meaningful here ?
I should use this result to prove $$\lim_{t\rightarrow \infty} (1+\frac a t)^t = e^a$$
 A: First of all, we recall that
$$\frac{d}{dx} \log(x) = \lim_{h \to 0} \frac{\log(x+h)-\log(x)}{h} = \frac{1}{x} \tag{1}$$
for any $x>0$. Now, as $\log(1)=0$, we find
$$\begin{align*} \lim_{t \to 0} \frac{\log(1+at)}{t} &= a \lim_{t \to 0} \frac{\log(1+at)-\log(1)}{at} \\ &= a \lim_{h \to 0} \frac{\log(1+h)-\log(1)}{h} \stackrel{(1)}{=} a \cdot 1. \end{align*}$$

Since $\mathbb{R} \ni x \mapsto \exp(x)$ and $(0,\infty) \ni x \mapsto \log(x)$ are continuous functions, the equality
$$\lim_{t \to \infty} \left(1+ \frac{a}{t} \right)^t = e^a$$
is equivalent to
$$\lim_{t \to \infty} \log \left[\left(1+ \frac{a}{t} \right)^t \right] = \log(e^a).$$
Using the standard calculus rules for the logarithm, we see that the latter is equivalent to
$$\lim_{t \to \infty} t \log \left(1+ \frac{a}{t} \right) = a.$$
Now the claim follows if we substitute $\frac{1}{t}$ by $h$ and consider the limit $h \to 0$ instead of $t \to \infty$.
A: Note that, if you plug in $t=0$ to the term $\dfrac{\log(1+at)}{t}$, you will get an indeterminate form $\dfrac{0}{0}$. Therefore, you need L'Hôpital's rule to evaluate
$$
\lim_{t\to0}\frac{\log(1+at)}{t}.
$$
I think the simplest way to obtain that limit is using Maclaurin series, but you have to use the definition of the differential quotient.
The differential quotient can also be written as
$$
\frac{f(x+h)-f(x)}{h},\quad\frac{f(t+h)-f(t)}{h},\quad\frac{f(t+\Delta t)-f(t)}{\Delta t},\quad\frac{f(z+\Delta z)-f(z)}{\Delta z},\text{ etc.}
$$
The symbol in the numerator part usually depends on the variable in the function and the symbol in the denominator part is arbitrarily chosen. In this case, let
$$
f(t)=\log(1+at)
$$
and you will have
$$
f(t+h)=\log(1+a(t+h))=\log(1+at+ah).
$$
Now, apply the L'Hôpital's rule and the definition of the differential quotient to obtain the limit equal to $a$. I will leave it this part to you.
After you successfully proving
$$
\lim_{t\to0}\frac{\log(1+at)}{t}=a,
$$
you can use it to show
$$
\lim_{t\to\infty}\left(1+\frac{a}{t}\right)^t=e^a.
$$
Rewrite
$$
\begin{align}
\lim_{t\to0}\frac{\log(1+at)}{t}&=a\\
\lim_{t\to0}\log(1+at)^\frac{1}{t}&=a\\
\lim_{t\to0}(1+at)^\frac{1}{t}&=e^a,
\end{align}
$$
then let $x=\dfrac{1}{t}\;\Rightarrow\;t=\dfrac{1}{x}$. As $t\to0$ implies $x\to\infty$. Hence,
$$
\begin{align}
\lim_{t\to0}(1+at)^\frac{1}{t}&=\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x.
\end{align}
$$
Now, considering the definition
$$
e=\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x.
$$
I think you can handle the rest, can't you? I hope this help, just let me know the rest of your work by commenting below.
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