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

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?

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

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

so:

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

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

Use L'Hôpital's rule

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

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

• You should have $\dfrac{d}{dx} (1-e^{6x}) = -6e^{6x}$, or else $d(1-e^{6x}) = -6e^{6x}\,dx$. But certainly not $(1-e^{6x})\,dx$. Apr 6, 2013 at 17:38
• Fixed it, careless today. Apr 6, 2013 at 17:57
• You haven't entirely fixed it yet. You've got $y/dx$ rather than $dy/dx$. Apr 6, 2013 at 18:02
• While first answer is more 'elegant', this is more general. You both have my +1. Apr 6, 2013 at 22:05
• You've a sign error in the last line. May 19, 2013 at 4:27

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

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