How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$$
I suppose in the second I may not take into account arctan and sin as sinx approximately equals x

Let us consider $$A= \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3}$$ and use Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$1+\sin(x)=1+x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^5\right)$$ $$1+\tan(x)=1+x+\frac{x^3}{3}+O\left(x^5\right)$$ Now, using the generalized binomial theorem $$\sqrt{1 + \sin(x)}=1+\frac{x}{2}-\frac{x^2}{8}-\frac{x^3}{48}+\frac{x^4}{384}+O\left(
   x^5\right)$$ $$\sqrt{1 + \tan(x)}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{11 x^3}{48}-\frac{47 x^4}{384}+O\left(x^5\right)$$ $$\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}=\frac{x^3}{4}-\frac{x^4}{8}+O\left(x^5\right)$$ $$A=\frac{1}{4}-\frac{x}{8}+O\left(x^2\right)$$ which not only shows the limit but also how it is approached.
