<s>[INST] Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\\&=\lim\limits_{x \to 0} \frac{\frac{\tan x-\sin x}{x^3}}{\sqrt{1+\tan x}+\sqrt{1+\sin x}}\end{align}$$
I can't proceed anymore from here.
 [/INST] $$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
=\frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}.$$
From Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $\tan x=x+x^3/3+O(x^5)$.
Therefore
$$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
\sim\frac{x^3/2}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}$$
as $x\to0$, and so the limit is $1/4$.
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