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?
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5 Answers
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Hint: $$1-e^{6x}=\left(1-e^{3x}\right)\left(1+e^{3x}\right)$$
answered Apr 6, 2013 at 17:27
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$$1-e^{6x}=(1-e^{3x})(1+e^{3x})$$
so:
$$\lim_{x\to 0}\frac{1-e^{6x}}{1-e^{3x}}= \lim_{x\to0} (1+e^{3x})=2$$
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$$\lim_{x \to 0}{\frac{1-e^{6x}}{1-e^{3x}}}=\lim_{x \to 0}{\frac{(1-e^{3x})(1+e^{3x})}{1-e^{3x}}}=1+1=2$$
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Use L'Hôpital's rule
$$\frac{d(1-e^{6x})}{dx}=-6e^{6x}$$ $$\frac{d(1-e^{3x})}{dx} = -3e^{3x}$$
$$\lim_{x\to 0}\frac{-6e^{6x}}{3e^{3x}} = 2$$
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1$\begingroup$ You should have $\dfrac{d}{dx} (1-e^{6x}) = -6e^{6x}$, or else $d(1-e^{6x}) = -6e^{6x}\,dx$. But certainly not $(1-e^{6x})\,dx$. $\endgroup$ Apr 6, 2013 at 17:38
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$\begingroup$ You haven't entirely fixed it yet. You've got $y/dx$ rather than $dy/dx$. $\endgroup$ Apr 6, 2013 at 18:02
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$\begingroup$ While first answer is more 'elegant', this is more general. You both have my +1. $\endgroup$ Apr 6, 2013 at 22:05
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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
$$\frac{-6x}{-3x} = 2$$
