# 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.

• $\frac{\tan(x)-\sin(x)}{x^3}=\frac{\sin(x)}{x}\frac{1-\cos(x)}{x^2}\frac{1}{\cos(x)}=\frac{\sin(x)}{x}\frac{\cos(0)-\cos(x)}{x^2}\frac{1}{\cos(x)}=\frac{\sin(x)}{x}\frac{2\sin(x/2)^2}{x^2}\frac{1}{\cos(x)}$ – logarithm May 2 at 6:47
• @logarithm Thanks! but I think this is easier: – firia2000 May 2 at 8:11
• $\frac{1-cosx}{x^2}=\frac{(1-cosx)(1+cosx)}{x^2(1+cosx)}=\frac{sin^2x}{x^2(1+cosx)}=(\frac{sinx}{x})^2\frac{1}{1+cosx}$ – firia2000 May 2 at 8:11

$$\frac{\tan x- \sin x}{x^3}=\frac{1}{ \cos x} \cdot \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2}.$$
$$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3} =\frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}.$$ 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 $$1/4$$.
Hint: $$\tan x- \sin x=\frac{x^3}{3}-\frac{x^3}{3!}+......$$