For Julian I

We can use Maclaurin series as follows:

$\begin{aligned} \lim_{h \to 0} {\frac{e^{2h}-1}{h}} & = \lim_{h \to 0} {\frac{\left(1 + 2h + \frac{(2h)^2}{2!} + \frac{(2h)^3}{3!} + ... \right) -1}{h}} \\ & = \lim_{h \to 0} \left( 2 + \frac{2^2h}{2!} + \frac{2^3h^2}{3!} + \frac{2^4h^3}{4!} + ... \right) \\ & = \boxed{2} \end{aligned}$

By the difference of two squares, $e^{2h}-1=(e^h+1)(e^h-1)$ , so $\displaystyle\lim_{h\to0} \dfrac{e^{2h}-1}{h}=2\times\lim_{h\to0}\dfrac{e^h-1}{h}$ , because $e^0+1=2$ . But, remember your calculus 1 classes? The derivative of a exponential function is, by definition, $\dfrac{d}{dx}n^x=\displaystyle\lim_{h\to0}\dfrac{n^{x+h}-n^x}{h}=n^x\times\lim_{h\to0}\dfrac{n^h-1}{h}$ , and, by definition, $e$ is the number which $\displaystyle\lim_{h\to0}\dfrac{n^h-1}{h}=1$ , in order to find the exponencial function which is its own derivative. So, $2\times\displaystyle\lim_{h\to0}\dfrac{e^h-1}{h}=2\times1=\boxed{2}$ .

Let 2h=t,as h->0 t->0 then it becomes 2lim e^t -1/t = 2 as t->0

Cheat, use a sci calculator and set h = to a small value, say .001, and calculate directly.

It is known that $\displaystyle\lim_{h\rightarrow0}\dfrac{e^h-1}{h}=1$ (don't trust me? Check a proof for $e^x$ being its own derivative). Now, $\displaystyle\lim_{h\rightarrow0}\dfrac{e^{2h}-1}{h}=\lim_{h\rightarrow0}\dfrac{e^h-1}{h}\cdot(e^h+1)=\lim_{h\rightarrow0}\dfrac{e^h-1}{h}\cdot\lim_{h\rightarrow0}(e^h+1)=1\cdot(1+1)=2$ .

The Taylor expansion of $e^x$ about $x=0$ is

$\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$

The first two terms of the expansion of $e^{2h}$ are $1+2h$ which gives $\frac{1+2h-1}{h}=\frac{2h}{h}=\boxed{2}$