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I am having trouble solving this limit. I tried applying L'Hôpital's rule but I got $\frac{0}{0}$.

$$\lim_{x\to0} {\frac{\frac{1}{1+x^3} + \frac{1}{3}\log{\left(1+3x^3\right)}-1}{2\sin{\left(3x^2\right)}-3\arctan{\left(2x^2\right)}}}$$

I would appreciate any hints in the right direction. Thanks in advance for your help.

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  • $\begingroup$ Have you tried to use L'Hopital's rule more than one time ? Like iterating unless you get the result you are searching for ? $\endgroup$
    – AlienRem
    May 28, 2015 at 11:31
  • $\begingroup$ Yes I have tried twice, should I try even more?? $\endgroup$
    – user244101
    May 28, 2015 at 11:32
  • $\begingroup$ I would try until I got a result and then maybe try with other methods $\endgroup$
    – AlienRem
    May 28, 2015 at 11:58

2 Answers 2

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Recall some standard limits (which can be computed without using l'Hospital's rule ): \begin{eqnarray*} \lim_{u\rightarrow 0}\frac{\frac{1}{1+u}-1+u}{u^{2}} &=&1 \\ \lim_{u\rightarrow 0}\frac{\log (1+u)-u}{u^{2}} &=&-\frac{1}{2} \\ \lim_{u\rightarrow 0}\frac{\sin u-u}{u^{3}} &=&-\frac{1}{6} \\ \lim_{u\rightarrow 0}\frac{\arctan u-u}{u^{3}} &=&-\frac{1}{3}. \end{eqnarray*} Now one can re-write the original expression using those of the standard limits above as follows \begin{eqnarray*} {\frac{\frac{1}{1+x^{3}}+\frac{1}{3}\log {\left( 1+3x^{3}\right) }-1}{2\sin {% \left( 3x^{2}\right) }-3\arctan {\left( 2x^{2}\right) }}} &{=}&\frac{\left( \frac{1}{1+x^{3}}-1+x^{3}\right) +\left( \frac{1}{3}\log {\left( 1+3x^{3}\right) -x}^{3}\right) }{\left( 2\sin {\left( 3x^{2}\right) -6x}% ^{2}\right) -\left( 3\arctan {\left( 2x^{2}\right) -6x}^{2}\right) } \\ &=&\frac{\left( \frac{\frac{1}{1+(x^{3})}-1+(x^{3})}{(x^{3})^{2}}\right) +3\times \left( \frac{\left( \log {\left( 1+\left( 3x^{3}\right) \right) -}% \left( {3x}^{3}\right) \right) }{(3x^{3})^{2}}\right) }{2\times 3^{3}\left( \frac{\sin {\left( 3x^{2}\right) -(3x}^{2})}{(3x^{2})^{3}}\right) -3\times 2^{3}\left( \frac{\arctan {\left( 2x^{2}\right) -(2x}^{2})}{\left( 2x^{2}\right) ^{3}}\right) } \end{eqnarray*} It follows that \begin{equation*} \lim_{x\rightarrow 0}\left( \frac{\frac{1}{1+x^{3}}+\frac{1}{3}\log {\left( 1+3x^{3}\right) }-1}{2\sin {\left( 3x^{2}\right) }-3\arctan {\left( 2x^{2}\right) }}\right) =\frac{\left( 1\right) +3\times \left( -\frac{1}{2}% \right) }{2\times 3^{3}\left( -\frac{1}{6}\right) -3\times 2^{3}\left( -% \frac{1}{3}\right) }=\frac{1}{2}. \end{equation*}

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Hint: $$\frac{1}{1+x^3} = 1 - x^{3} + x^{6} - x^{9} + O(x^{12}) \,\,\, \text{and} \,\,\, \log (1 + 3x^3) = 3x^3 - \frac{9x^6}{2} - 9x^9 + O(x^{12})$$

$$\sin (3x^2) = 3x^2 - \frac{9x^6}{2} + O(x^{10}) \,\,\, \text{and} \,\,\, \arctan (2x^2) = 2x^2 - \frac{8x^6}{3} + O(x^{10}) $$

Can you take it from here?

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  • $\begingroup$ what is $O(x^{12})$? $\endgroup$
    – MonK
    May 28, 2015 at 11:35
  • $\begingroup$ Check Big O notation. $\endgroup$ May 28, 2015 at 11:37
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    $\begingroup$ It means that there are terms omitted, terms that go to zero as $x\rightarrow 0$ at least as fast as $x^{12}$. It also implies that those terms are negligible compared to the terms not omitted. $\endgroup$ May 28, 2015 at 11:37
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    $\begingroup$ @grdgfgr Not strictly faster, but at least as fast as $x^{12}$. $\endgroup$
    – naslundx
    May 28, 2015 at 11:40
  • $\begingroup$ @Aaron Maroja, yes, I think I get your idea, really appreciate it. $\endgroup$
    – user244101
    May 28, 2015 at 20:30

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