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

Question

$$\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^{3}}$$

How would I find the limit with out using conjugate nor L'Hospital's rule.

Answer 1

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 $x \to 0$,

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

Answer 2

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.