$$ \lim_{x \rightarrow 0} \frac{e^{x\sin(x)}+e^{x\sin(2x)}-2}{x\ln(1+x)} $$
Hello I've recently started learning calculus and I got curious about the solution of the problem above.
I was about to write down my solution, but I soon realized that my solution was totally wrong while writing--so I erased all and changed title.
Please let me know how to solve this: steps or rules that I should use. It is quite alright to not be a long, detailed solution but to be short, concise instructions.
Thanks!
$\displaystyle\lim_{x \to 0} \frac{e^{x\sin(x)}+e^{x\sin(2x)}-2}{x\ln(1+x)}=\lim_{x \to 0}\frac{e^{x \sin x}-1}{x \ln(1+x)}+\lim_{x \to 0}\frac{e^{x \sin 2x}-1}{x \ln(1+x)}$
Now you can use some known limit to evaluate $\lim_{x \to 0}\frac{e^{x \sin x}-1}{x \ln(1+x)}$ and $\lim_{x \to 0}\frac{e^{x \sin 2x}-1}{x \ln(1+x)}$.
You know that $\displaystyle \lim_{y \to 0}\frac{e^y-1}{y}=1$ (l'Hospital, Taylor series or others) and $\lim_{x \to 0}x \sin x=0$, so $$\lim_{x \to 0} \frac{e^{x\sin x}-1}{x \sin x}=1$$
Next you can use two known limits $\displaystyle \lim_{y \to 0}\frac{\sin y}{y}=$ and $1=\displaystyle \lim_{y \to 0}\frac{\ln(1+y)}{y}=1$ (also l'Hospital, Taylor series or others) $\displaystyle \lim_{x \to 0} \frac{e^{x\sin x}-1}{x \sin x}=\lim_{x \to 0} \frac{e^{x\sin x}-1}{x \sin x}\frac{x\ln(1+x)}{x\ln(1+x)}=\lim_{x \to 0} \frac{e^{x\sin x}-1}{x \ln (1+x)}\lim_{x \to 0}\frac{x \ln(1+x)}{x\sin x}=\lim_{x \to 0} \frac{e^{x\sin x}-1}{x \ln (1+x)} \cdot \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0}\frac{\ln(1+x)}{x}=\lim_{x \to 0} \frac{e^{x\sin x}-1}{x \ln (1+x)}$
The same with the second limit, but with $$\lim_{x \to 0} \frac{e^{x\sin 2x}-1}{x \sin 2x}=1$$ and $\lim_{y \to 0}\frac{\sin 2y}{2y}=1$, so $\lim_{y \to 0}\frac{\sin 2y}{y}=2$
Hint
You have an expression which is $$f(x)=\frac{u(x)}{v(x)}$$ and $u(0)=0$,$v(0)=0$ which makes the ratio undertermined. You probably learnt L'Hospital rule which uses the fact that, in such a case, the limit of $f(x)$ is the same as the limit of $\frac{u'(x)}{v'(x)}$. If both quantities are not zero, you are done; but, if both are again equal to zero, you must continue the process using the fact that the limit will be the same as the limit of $\frac{u''(x)}{v''(x)}$.
The other solution would consist in the composition of Taylor series built at $x=0$ but I do not know if you already heard of that.
$$\lim_{x \rightarrow 0} (e^{x\sin(x)}+e^{x\sin(2x)}-2)=\lim_{x \rightarrow 0}3x^2$$
$$\lim_{x \rightarrow 0} (x\ln(1+x))=\lim_{x \rightarrow 0} x^2$$
So
$$\lim_{x \rightarrow 0} \frac{e^{x\sin(x)}+e^{x\sin(2x)}-2}{x\ln(1+x)}=3$$
