It's a while I'm trying to verify this limit: $$\lim_{x\rightarrow 0^+} \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0,$$ this is my attempt:
Since $x \rightarrow 0$ I can do the substitution: $$\sqrt{1+x^2}=1+o(1)$$ $$\sin{x}=x(1+o(1))$$ and I obtain: $$\frac{\log(x+1+o(1))-x(1+o(1))}{(2^{x^2}-1)(\sin (e^{x^2}-1))} = 0$$
Now I do these substitutions for $y \rightarrow 0$ $$\log (y+1)= y(1+o(1))$$ $$a^y-1 = y(\log a + o(1))$$ $$\sin y = y(1 + o(1))$$ and I obtain: $$\frac{(x+o(1))(1+o(1))-x(1+o(1))}{x^2(\log 2+o(1))x^2(1+o(1))}=\frac{x+o(1)-x+o(x)}{(x^2 \log 2 + o(x^2))(x^2+o(x^2))}=\frac{o(1)}{x^4 \log 2 + o(x^4)}.$$ But the limit with $x \rightarrow 0$ of this makes infinite, not $0$ as it should.
Where is my error? Thank you.
You have \begin{align} \sqrt{1+x^2}&=1+\frac{x^2}2-\frac{x^4}8+o(x^6)\\ \ \\ \log(1+x)&=x-\frac{x^2}2+\frac{x^3}3+o(x^4)\\ \ \\ \sin x&=x-x^3/6+o(x^5)\\ \ \\ 2^x&=1+x\log2+o(x^2)\\ \ \\ e^x&=1+x+o(x^2) \end{align} Then \begin{align} \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))} &=\frac{\log(1+x+x^2/2-x^4/8+o(x^6))-(x-x^3/6+o(x^5))}{(x^2\log2+o(x^4))(\sin(x^2+o(x^4)))}\\ \ \\ &=\frac{\log(x+\sqrt{1+x^2})-\sin x}{(x^2\log2+o(x^4))(x^2+o(x^4))}\\ \ \\ &=\frac{\log(x+\sqrt{1+x^2})-\sin x}{o(x^4)}. \end{align} So the denominator is clearly $o(x^4)$. Let us work on the numerator: using Wolphram Alpha, \begin{align} \log(x+\sqrt{1+x^2})&=\log(1+x+\frac{x^2}2-\frac{x^4}8+o(x^6))\\ \ \\ &=x+\frac{x^2}2-\frac{x^4}8-\frac{(x+\frac{x^2}2-\frac{x^4}8)^2}2 +\frac{(x+\frac{x^2}2-\frac{x^4}8)^3}3+o(x^6)\\ \ \\ &=x-\frac{x^3}6+o(x^5). \end{align} Then $$ \log(x+\sqrt{1+x^2})-\sin x=x-\frac{x^3}6+o(x^5)-\left(x-\frac{x^3}6+o(x^5)\right)=o(x^5). $$ Thus $$ \frac{\log(x+\sqrt{1+x^2})-\sin x}{(2^{x^2}-1)(\sin (e^{x^2}-1))}=\frac{o(x^5)}{o(x^4)}=o(x). $$