Limits without L'Hospital

Question

Evaluate:

$$\lim_{h \rightarrow 0} \frac{e^{2h}-1}{h}$$

Now one way would be using the Maclaurin expansion for $e^{2x}$

However, can we solve it using the definition of the derivative (perhaps considering $f(x)=e^x$)? Many thanks for your help! $$$$ EDIT: I forgot to mention to please not use L'Hopital's Rule. Using it, the problem becomes trivial and loses all chances of getting a beautiful solution.

Answer 1

Noting $e^{2h}-1=(e^h-1)(e^h+1)$ and recalling a notable limit does the job.

Answer 2

$$\lim_{h \rightarrow 0} \frac{e^{2h}-1}{h} = 2 \lim_{h \to 0}\frac{e^{2h}-1}{h} = 2\lim_{x \to 0}\frac{e^x-1}{x} = 2\cdot 1 = 2,$$ where I made the substituition $x = 2h$ just to make things easier for you to visualize. I used one of the fundamental limits along with the fact that $h \to 0 \iff x \to 0$.

Answer 3

Note that, with $f(x) = e^x$, we have $\lim \limits_{h \to 0} \dfrac{e^h - 1}{h} = f'(0) = 1$.

Therefore, $\lim \limits_{h \to 0} \frac{e^{2h} - 1}{h} = \lim \limits_{h \to 0} \frac{(e^h - 1)}{h}(e^h + 1) = 1 \cdot (1 + 1) = \boxed 2$.

Answer 4

Using $e^{2h}-1=(e^h-1)(e^h+1)$, we get

\begin{align*} \lim_{h \rightarrow 0} \frac{e^{2h}-1}{h} &= 2 \lim_{h \to 0}\frac{ (e^h-1)(e^h+1)}{h} \\&= \lim_{h \to 0}(e^h+1)\lim_{h \to 0}\frac{e^h-1}{h} \\&= 2\lim_{h \to 0}\frac{e^h-1}{h} \\&=2\lim_{h \to 0}\frac{e^{0+h}-e^0}{h} \\&=2f'(0)=2e^0=2\cdot1=2\end{align*}

with $f(x)=e^x$ of course.