I have this limit: $$\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}$$
And I try this \begin{align}\lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}} & = \lim_{x\to 0^+} \frac{\sqrt{1+2x} - e^x}{x\arctan{x}}\frac{\sqrt{1+2x} + e^x}{\sqrt{1+2x} + e^x} \\ & \stackrel{(*)}= \lim_{x\to 0^+} \frac{1+2x - e^x}{x^2(\sqrt{1+2x}+e^x)} \end{align} $(*)$: $\arctan x\simeq x$
But now I'm blocked... Any advice?
Considering $$A=\frac{\sqrt{1+2x} - e^x}{x\,\tan^{-1}(x)}$$ there are two possibilities.
The first one is to use L'Hôpital's rule just as Workaholic commented, setting $$u=\sqrt{1+2x} - e^x\qquad , \qquad v=x\,\tan^{-1}(x)$$ and you will probaly need to apply it more than once.
The second one would involve Taylor expansions $$\sqrt{1+2x}=1+x-\frac{x^2}{2}+\frac{x^3}{2}+O\left(x^4\right)$$ $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)$$ $$\tan^{-1}(x)=x-\frac{x^3}{3}+O\left(x^4\right)$$ which make $$A=\frac{-x^2+\frac{x^3}{3}+O\left(x^4\right)}{x^2-\frac{x^4}{3}+O\left(x^5\right)}$$ Now, long division to get $$A=-1+\frac{x}{3}+O\left(x^2\right)$$ which shows the limit and also how it is approached.
