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My friend passed me a limit that (in his opinion) is resovable using only derivatives or similar method. Here is the limit: $$\lim_{x\to0} \frac{\ln(1+x+x^2) - (e^x-1)}{x\sin(x)}$$ I tried to solve it using basic limits (the limit form clearly shows parts of notable limits) but my solution is $1$ that is wrong. Can someone solve it avoiding derivative methods or similar methods? Or my friend is right? Sorry for my english and thank you in advance (if what I've written is not enough please advice me)

I'm supposed to solve it knowing only basic limits and algebric manipulation nothing more. Just as someone that started to do limits in his first steps

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  • $\begingroup$ Does not using derivatives also mean not using polynomial approximations, which of course have their basis in the derivatives of the relevant functions? $\endgroup$
    – Simon S
    Commented Nov 1, 2014 at 14:53
  • $\begingroup$ OP said "using only derivatives or similar methods". But I guess Taylor expansions would be to much. $\endgroup$ Commented Nov 1, 2014 at 14:55
  • $\begingroup$ I'm supposed to solve it knowing only basic limits and algebric manipulation nothing more. Just as someone that started to do limits in his first steps $\endgroup$
    – Dipok
    Commented Nov 1, 2014 at 14:56
  • $\begingroup$ @Dipok How do you know that e.g. $\lim_{x\to0}\frac{\sin x}x=1$ then? $\endgroup$ Commented Nov 1, 2014 at 15:13
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    $\begingroup$ @dipok: When you say you want to "only" use basic limits and algebraic manipulation, what do you mean exactly? How do you define $\ln$, $e^x$ and $\sin x$? $\endgroup$ Commented Nov 1, 2014 at 15:19

3 Answers 3

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I don't think you can prove it without knowing that $$\ln(1+x)=x+x^2/2+o(x^2), e^x=1+x+x^2/2+o(x^2)\text{ and }\sin x=x+o(x),$$ This is well known and you actually know special cases of this, such as \begin{align*}&\lim_{x\to0}\frac{\ln(1+x)}x=1\iff \lim_{x\to0}\frac{\ln(1+x)-x}x=0\iff \ln(1+x)=x+o(x)\\&\lim_{x\to0}\frac{e^x-1}x=1\iff \lim_{x\to0}\frac{e^x-1-x}x=0\iff e^x=1+x+o(x)\end{align*} If you know what I wrote at the beginning, you can easily compute the limit: \begin{align*}&\lim_{x\to0}\frac{\ln(1+x+x^2)-(e^x-1)}{x\sin x}=\\&\lim_{x\to0}\frac{(x+x^2)-(x+x^2)^2/2+o(x^2)-(1+x+x^2/2+o(x^2)-1)}{x(x+o(x))}=\\&\lim_{x\to0}\frac{x+x^2-x^2/2+o(x^2)-x-x^2/2+o(x^2)}{x^2+o(x^2)}=\lim_{x\to0}\frac{o(1)}{1+o(1)}=0\end{align*}

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  • $\begingroup$ When you are using Taylor polynomials, you are using derivatives. I would not accept this solution. Method of Taylor polynomials is even stronger method than LHospital rule. $\endgroup$
    – Joseph
    Commented Nov 1, 2014 at 16:12
  • $\begingroup$ @Joseph Not necessarily. If the series definitions of the functions are used, then you don't need anything. $\endgroup$ Commented Nov 1, 2014 at 18:06
  • $\begingroup$ Yes. If you define functions as series, then your solution is elementary. $\endgroup$
    – Joseph
    Commented Nov 1, 2014 at 18:10
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$$\ln(1+x+x^2)\sim_0 x+x^2$$ and $$\sin x\sim_0 x$$

Moreover $$e^x=1+x+\frac{x^2}{2}+x^2\varepsilon(x)$$ if $x\to 0$ where $\lim_{x\to 0}\varepsilon(x)=0$

then $$...=\lim_{x\to 0}\frac{x+x^2-(1+x+\frac{x^2}{2}+x^2\varepsilon(x)-1)}{x^2}=\lim_{x\to 0}\varepsilon(x)=0$$

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    $\begingroup$ if the person who downvote could explain, it would be more constructive. $\endgroup$
    – idm
    Commented Nov 1, 2014 at 15:38
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You can answer the question using standard definitions of Taylor series expansions:

$\ln \left(1+x\right) = x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3} - \ldots$ ,

$e^{x} = 1+x+\dfrac{x^{2}}{2!} + \ldots$ and $\sin \left (x\right) = x - \dfrac{x^{3}}{3!} + \ldots$

Now the only expression we need to worry about expanding is the $\ln$ term $$\ln \left(1+x+x^{2}\right) = \left(x+x^{2}\right) - \dfrac{\left(x+x^{2}\right)^{2}}{2} + \dfrac {\left(x+x^{2}\right)^{3}}{3} - \ldots$$

$$= x + x^{2} - \dfrac{x^{2}}{2}\left(1+x\right)^{2} + \dfrac{x^{3}}{3}\left(1+x\right)^{3} + \ldots$$ which after some expanding and collecting terms gives:

$$\ln \left(1+x+x^{2}\right) = x + \dfrac{1}{2}x^{2} - \dfrac {2}{3}x^{3} + \ldots$$

We substitute this series and the other series into the expression concerned:

$$\lim_{x\rightarrow 0} \dfrac {\ln \left(1+x+x^{2}\right) - e^{x} + 1}{x\sin \left (x\right)} = \lim_{x\rightarrow 0}\dfrac{x+\dfrac{1}{2}x^{2}-\dfrac{2}{3}x^{3}-\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^{3}}{3!}+\ldots\right) + 1}{x\left(x-\dfrac{x^{3}}{6}+\ldots\right)}$$

$$ = \lim_{x\rightarrow 0} \dfrac{x+\dfrac{1}{2}x^{2}-\dfrac{2}{3}x^{3}- x - \dfrac{x^{2}}{2} - \dfrac{x^{3}}{6}-\ldots}{x^{2}-\dfrac{x^{4}}{6}+\ldots}$$

$$ = \lim_{x\rightarrow 0} \dfrac{- \dfrac{5}{6}x^{3} + \ldots}{x^{2}-\dfrac{x^{4}}{6}+\ldots}$$

Divide through by the $x^{2}$ in the fraction to get:

$$\lim_{x\rightarrow 0} \dfrac {\ln \left(1+x+x^{2}\right) - e^{x} + 1}{x\sin \left (x\right)} = \lim_{x\rightarrow 0} \dfrac{ - \dfrac{5}{6}x + \ldots}{1 - \dfrac{x^{2}}{6}+\ldots} = 0$$

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  • $\begingroup$ So we HAVE to use Taylor expansion to solve this limit? $\endgroup$
    – Dipok
    Commented Nov 1, 2014 at 15:58

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