<s>[INST] 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
 [/INST] The second you've got down pat.
$$\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = \lim_{x \to 0} \frac{\arctan 2x}{x}\frac{x}{\sin(2\pi x)} = \frac{1}{\pi}$$
The first is a little trickier.  Use the binomial theorem to get...
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} = \lim_{x \to 0} \frac{\tan x - \sin x}{2x^3} = \lim_{x \to 0} \frac{\sin x }{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{2\cos x} = \frac{1}{4}$$
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