Limit solving without using L'Hopital rule

Question

How can I solve this limit

$$ \lim_{x \to 0}\frac{e^{x+1}-e}{3x} $$

without using L'Hopital's rule?

I know this is true: $$ \lim_{x \to 0}\frac{e^{x}-1}{x} = 1 $$

So i belive we have to use this in anyway possible.

Answer 1

$$\lim_{x\to 0} \dfrac{e^{x+1} - e }{3x} = \dfrac{e}{3} \lim_{x\to 0}\dfrac{e^x - 1}{x} = \dfrac{e}{3}$$

Answer 2

Hint You can use the infinite series expansion of $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots$

Answer 3

$$\ \lim_{x\to0}\frac{e^{x+1}-e}{3x}=\lim_{x\to0}\frac{e\cdot e^x-e}{3x}\lim_{x\to0}\frac{e}{3}\frac{e^x-1}{x}=\frac{e}{3}$$

Remember that, if $\ a\in \mathbb R$ then $\ a^{x+y}=a^x\cdot a^y$

Answer 4

HINT

$\displaystyle \lim_{x \to x_0} \frac{e^x - e^{x_0}}{x-x_0}=(e^x)_{x=x_0}'=e^{x_0}$

Then I would say

$\displaystyle \lim_{x \to 0}\frac{e^{x+1}-e}{3x}=\frac{e}{3}\cdot \lim_{x \to 0} \frac{e^x-e^0}{x-0}=\frac{e}{3}\cdot (e^x)'_{x=0}=\frac{e}{3}$