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

Hint: Multiply the top and bottom of the first limit in order to rewrite it as
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3 (\sqrt{1+\tan(x)} + \sqrt{1 + \sin(x)})} = 
\frac 12 \lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3}
$$
Then we have
$$
\lim_{x \to 0} \frac{\tan(x) - \sin(x)}{x^3} = 
\lim_{x \to 0} \frac 1{\cos(x)}\frac{\sin(x)}{x} \frac{(1 - \cos(x))}{x^2} = \\
\lim_{x \to 0} \frac{1 - \cos(x)}{x^2}
$$
now apply a similar trick.

For the second, we can write
$$
\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = 
\lim_{x \to 0} \frac{\arctan 2x}{\sin(2 \pi x)} = \\
\lim_{x \to 0} \frac{1}{2\pi}\frac{\arctan 2x}{x}
\cdot \frac{2 \pi x}{\sin (2 \pi x)} =\\
\frac{1}{2\pi}\lim_{x \to 0} \frac{\arctan (2(0+x)) - \arctan(2(0))}{x}
$$
you might find it easier to find that second limit if you replace $x$ with an $h$ or a $\Delta x$.  Alternatively, with $\theta = \arctan(2x)$, we have
$$
\lim_{x \to 0} \frac{\arctan(2x)}{x} = \lim_{\theta \to 0} \frac{\theta}{\tan(\theta/2)} = 
2\lim_{\theta \to 0}
\cos (\theta/2)\frac{\theta/2}{\sin(\theta/2)}
$$
