1

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.

João Silva
  • 47
  • 1
    Are you familiar with the concept of derivative? How do you define exponential function? Please add such context to your question. –  Feb 08 '15 at 16:11
  • I am familiar with that concept but the exercise must not be resolved that way. I believe we have to use this $$ \lim_{x \to 0}\frac{e^{x}-1}{x} = 1 $$ – João Silva Feb 08 '15 at 16:14
  • What do you mean, "that way"? The equality you wrote in your comment is simply the derivative of $e^x$ when $x=0$. – Andrés E. Caicedo Feb 08 '15 at 16:16
  • @JoãoSilva if you already know that $$\lim_{x \to 0}\frac{e^{x}-1}{x}$$ then factor $\frac{e}{3}$ out in front of the limit. – graydad Feb 08 '15 at 16:18
  • My doubt is how you take out the +1 on e^x+1. – João Silva Feb 08 '15 at 16:22

4 Answers4

2

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

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

Mosk
  • 1,630
1

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

Samrat Mukhopadhyay
  • 16,277
1

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

abel
  • 29,170
0

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

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

georg
  • 2,749