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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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    I am sure you tried *something*. Can you show your attempt? To apply L'Hospital's rule (after checking that it is indeed applicable for this problem), you will need to find the derivatives of the numerator and the denominator. Could you find them? It's important for you to show your work.2011-10-17
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    That is what I can not do I have tried raising everything by ln and then I am not sure what to do after that, I have tried raising everything to en exponent ln but that didn't look right and I have also tried factoring out an exponent x but that definitely didn't work.2011-10-17
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    The top and bottom goes to zero. Use the product rule to differentiate the top part. To differentiate the bottom part, use this trick $y=3^x$, $x=\frac{ln(y)}{ln(3)}$, then differentiate that to get $\frac{dx}{dy}$ the trick is then to flip it and then sub y in it.2011-10-17
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    I don't understand the part with the lny/ln3=x.2011-10-17
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    You take the $log_3$ in both sides. Then you need to change that into $log_e$ which is ln. You can just also remember that $y=a^x$ differentiates to $ln(a) \times a^x$ if you can't follow that. It's called implicit differentiation.2011-10-17
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    I know about implicit differentiation I just don't understand how log3 can turn into ln.2011-10-17
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    @Jordan: You either need to know the formula for differentiating an arbitrary exponential, $$\frac{d}{dx}a^x = \ln(a)a^x,$$where $a$ is any real number greater than $0$, or you need to use the Change-of-Basis formula for logarithms, $$\log_a(x) = \frac{\ln(x)}{\ln(a)},$$where $a$ is any real number greater than $0$ and different from $1$; or you need to do some basic algebra with logarithms: if $y=3^x$, then taking ln on both sides you get $\ln(y) = \ln(3^x) = x\ln(3)$ (last equality by the basic properties of logarithms). Now solve for $x$. Once again, your algebra skills hold you back.2011-10-18
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    @ArturoMagidin I know it isn't applicable here but why can't I factor out an x from the numerator and denominator which will then make it not an indeterminate form if I make the x/x into 1. I have seen this being an option after using hospitals rule but it will not give me a correct answer I always have to use the rule again. Also I am pretty sure I am going to take college algebra and trig over again although I didn't learn any algera in those classes, not sure why I would a second time. I basically just wasted a lot of money and about a year of college.2011-10-19
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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

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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 don't understand what is happening, $3^x$ equals that but then how is the derivative of that just ln3e should there be an x in the base?2011-10-17
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    [there](http://en.wikipedia.org/wiki/Chain_rule) you can find rules2011-10-17
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    I know what the chain rule is.2011-10-17
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    You need to look back at your basic algebra. He is using the power rules and the fact that $e^ln(3)=3$2011-10-17
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    Ok I have been doing that for 2 months now and now I am behind in my class. eln3=3? I don't even know what that means.2011-10-17
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    Jordan, I dont understand what's the problem, if you know chain rule and the fact that $3^x=e^{ln(3)x}$. It is all you need.2011-10-17
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    $e^{\ln 3}=3$ since $\ln 3$ is by definition the power that when $e$ is raised to gives $3$.2011-10-17
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    Jordan, do you know what is $e^x$ function?2011-10-17
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    Jordan it would be nice if you wrote what you knew. You said you have seen the chain rule, this lead me to believe you have seen exponential function. Plus you are asking a question about a very advanced rule beyond it.2011-10-17
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    Honestly I don't know what I know, I constantly forget everything I learn so it is hard to say. I am working on L'hospitals rule and optimization problems right now. I still don't quite understand what to do with this problem.2011-10-17
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    @Jordan: (1) You check that the limit is an indeterminate of form $\frac{0}{0}$ by pluggin in $x=0$ and noticing that you get $0$ on the numerator and $0$ on the denominator. (2) You check that both the numerator and the denominator are differentiable. (3) You decide to apply L'Hospital's Rule to compute the limit. (4) To apply L'Hospital's Rule, you differentiate the numerator $x3^x$, and separately, the denominator $3^x-1$. Then you have$$\lim_{x\to 0}\frac{x3^x}{3^x-1} = \lim_{x\to 0}\frac{(x3^x)'}{(3^x-1)'}.$$ (5) You do this new limit (which can be done by simply plugging in $x=0$).2011-10-18
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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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    I do not understand what you mean by second limit/2012-06-08
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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