How do you find the limit of $\lim_{x \to 0} \frac{1-e^{6x}}{1-e^{3x}}$

Question

I would like to know how to calculate the limit for:

$$\lim_{x \to 0} \frac{1-e^{6x}}{1-e^{3x}}$$

I tried by factoring by $$\frac{1-e^{3x}}{1-e^{3x}}$$

I'm not sure if this is correct. Am I doing something wrong?

score 10 · Answer 1 · answered Apr 06 '13 at 17:27

10

Hint: $$1-e^{6x}=\left(1-e^{3x}\right)\left(1+e^{3x}\right)$$

answered Apr 06 '13 at 17:27

Cameron Buie

score 8 · Accepted Answer · edited Apr 06 '13 at 17:32

8

$$1-e^{6x}=(1-e^{3x})(1+e^{3x})$$

so:

$$\lim_{x\to 0}\frac{1-e^{6x}}{1-e^{3x}}= \lim_{x\to0} (1+e^{3x})=2$$

edited Apr 06 '13 at 17:32

answered Apr 06 '13 at 17:29

masmoudihoussem

score 7 · Answer 3 · answered Apr 06 '13 at 17:29

7

$$\lim_{x \to 0}{\frac{1-e^{6x}}{1-e^{3x}}}=\lim_{x \to 0}{\frac{(1-e^{3x})(1+e^{3x})}{1-e^{3x}}}=1+1=2$$

answered Apr 06 '13 at 17:29

Iuli

score 7 · Answer 4 · edited Apr 06 '13 at 19:43

7

Use L'Hôpital's rule

$$\frac{d(1-e^{6x})}{dx}=-6e^{6x}$$ $$\frac{d(1-e^{3x})}{dx} = -3e^{3x}$$

$$\lim_{x\to 0}\frac{-6e^{6x}}{3e^{3x}} = 2$$

edited Apr 06 '13 at 19:43

DonAntonio

answered Apr 06 '13 at 17:34

igumnov

1

You should have $\dfrac{d}{dx} (1-e^{6x}) = -6e^{6x}$, or else $d(1-e^{6x}) = -6e^{6x},dx$. But certainly not $(1-e^{6x}),dx$. – Michael Hardy Apr 06 '13 at 17:38
Fixed it, careless today. – igumnov Apr 06 '13 at 17:57
You haven't entirely fixed it yet. You've got $y/dx$ rather than $dy/dx$. – Michael Hardy Apr 06 '13 at 18:02
While first answer is more 'elegant', this is more general. You both have my +1. – Francisco Presencia Apr 06 '13 at 22:05
1

You've a sign error in the last line. – Cameron Buie May 19 '13 at 04:27

score 1 · Answer 5 · answered Apr 06 '13 at 21:35

1

A fancy way to do it (that is indeed fancy in this case but it's the only way to get out alive from a lot of other cases):

$$1 - e^{3x} \sim -3x$$ $$1 - e^{6x} \sim -6x$$

since you can substitute, you get

$$\frac{-6x}{-3x} = 2$$

answered Apr 06 '13 at 21:35

Ant

5 Answers5