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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. –  Srivatsan Oct 17 '11 at 22:08
    
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. –  user138246 Oct 17 '11 at 22:09
    
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. –  simplicity Oct 17 '11 at 22:16
    
I don't understand the part with the lny/ln3=x. –  user138246 Oct 17 '11 at 22:19
    
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. –  simplicity Oct 17 '11 at 22:23
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3 Answers

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? –  user138246 Oct 17 '11 at 22:31
    
there you can find rules –  Savinov Evgeny Oct 17 '11 at 22:36
    
I know what the chain rule is. –  user138246 Oct 17 '11 at 22:37
    
You need to look back at your basic algebra. He is using the power rules and the fact that $e^ln(3)=3$ –  simplicity Oct 17 '11 at 22:40
    
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. –  user138246 Oct 17 '11 at 22:43
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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/ –  user138246 Jun 8 '12 at 18:43
    
@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. –  Michael Hardy Jun 8 '12 at 20:48
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$$\frac{(x3^x)'}{(3^x-1)'}$$ $$\frac{3^x+x3^x\ln3}{3^x\ln3}$$ Now you can substitute 0. You will get $1/\ln3$.

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Please use $\LaTeX$, complete sentences, and standard spelling. –  Nate Eldredge Jan 8 '12 at 20:38
    
i am usin mobile phone in ethiopia.sorry i don hav access even i don know what u mean. –  albo Jan 9 '12 at 14:36
    
How is this possible? I thought that you have to raise everything to ln to get the derivative. –  user138246 Mar 30 '12 at 20:49
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