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I am trying to find $\lim_{x\to 0}\frac{x3^x}{3^x-1}.$

I don't really know where to start, I know I have done this many times but I can never remember what to do for this and there are no practice problems for this in my book.

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    @Jordan: You cannot factor $x$ from the denominator because the denominator is not$a$multiple of $x$. It is **not** true that if$a$function evaluate to $0$ at $0$ then it has a factor of $x$. That is only true for polynomials. The function $3^x-1$ is **not** a multiple of $x$, so you cannot "factor $x$. The only way to factor $x$ out of the denominator would be by including a division by $x$, $3^x-1 = x\left(\frac{3^x-1}{x}\right)$ which is *still* an indeterminate at $0$: you solved nothing.2011-10-19

2 Answers 2

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It helps you: $3^x=e^{\ln(3)x}$; (3^x)'=\ln(3)e^{\ln(3)x}=\ln(3)3^x

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    I am glad I am reviewing this thread, this is an example of math people being elitist and arrogant.2012-03-30
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$ \lim_{x\to0}\frac{x3^x}{3^x-1} = \left(\lim_{x\to0} 3^x \right)\left(\lim_{x\to0}\frac{x}{3^x-1}\right). $ L'Hôpital's rule does the second limit; the first is trivial.

Alternatively, you could say that the second limit is the reciprocal of $\lim\limits_{x\to 0}\dfrac{3^x - 1}{x}$, and that is $\left.\dfrac{d}{dx} 3^x \right|_{x=0}$.

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    @Jordan : On the first line of the answer, just after "$=$", you see two limits being multiplied. When I said "the second limit", I meant the second one of those.2012-06-08