Solving a limit without l'Hopital

Question

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

Answer 1

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*}

Answer 2

$$\ln(1+x+x^2)\sim_0 x+x^2$$ 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 $$...=\lim_{x\to 0}\frac{x+x^2-(1+x+\frac{x^2}{2}+x^2\varepsilon(x)-1)}{x^2}=\lim_{x\to 0}\varepsilon(x)=0$$

Answer 3

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:

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