How to obtain $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$ without L'Hospital's rule? 
$$\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$$ 

How would I find the limit with out using conjugate nor L'Hospital's rule.
 A: You may observe that, by the Taylor expansion, you have, as $x \to 0$,
$$
\begin{align}
\sin x&=x-\frac{x^3}6+O(x^5) \tag1
\\ \tan x&=x+\frac{x^3}3+O(x^5) \tag2
\end{align}
$$ and, as $u \to 0$, $$
\begin{align}
\sqrt{1+u}&=1+\frac{u}2-\frac{u^2}8+\frac{u^3}{16}+O(u^4). \tag3
\end{align}
$$ Thus
$$
\sqrt{1+\tan x}-\sqrt{1+\sin x}=\frac{x^3}4+O(x^4) \tag4
$$ giving, as $x \to 0$,

$$
\lim_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}} = \frac14.
$$

A: Rationalising, we have
\begin{align}
\dfrac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3} & = \dfrac{\tan(x)-\sin(x)}{x^3\left(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)}\right)}\\
& = \dfrac{\sin(x)}x \cdot \dfrac{1-\cos(x)}{x^2} \cdot \dfrac1{\cos(x)\left(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)}\right)}\\
& = \dfrac{\sin(x)}x \cdot \dfrac{2\sin^2(x/2)}{x^2} \cdot \dfrac1{\cos(x)\left(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)}\right)}
\end{align}
Hence,
$$\lim_{x \to 0}\dfrac{\sqrt{1+\tan(x)}-\sqrt{1+\sin(x)}}{x^3} = \lim_{x \to 0}\dfrac{\sin(x)}x \cdot \lim_{x \to 0}\dfrac{2\sin^2(x/2)}{x^2} \cdot \lim_{x \to 0}\dfrac1{\cos(x)\left(\sqrt{1+\tan(x)}+\sqrt{1+\sin(x)}\right)}$$
Trust you can finish it off from here.
