Fastest way to obtain limit of a function I'm tasked to find the limit of such a function:
$\lim_{x\to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3}$
My immediate instinct is to use L'Hospital's Rule to differentiate the numerator and denominator, and rinse and repeat until the denominator no longer contained $x$, which I did to arrive at an answer of $0$, but boy was the process extremely tedious and painful.
However, this process was way too painful. My other instinctive thought was to rationalise the function to obtain $\lim_{x\to 0} \frac{\tan(x) - \sin(x)}{x^3\bigl(\sqrt{1 + \tan(x)} + \sqrt{1 + \sin(x)}\bigr)}$ but this not only does not remove $x$ from the denominator, I have to apply the product rule which will no doubt complicate the process further.
Can someone advise how else I can go about solving this more efficiently?
 A: L'Hospital's rule is not the alpha and omega of limits computation!
From the second approach, you can use Taylor's formula at order $3$ and equivalence of functions:
$$\tan x-\sin x=x+\frac{x^3}3+o(x^3)-\Bigl(x-\frac{x^3}6+o(x^3)\Bigr)=\frac{x^3}2+o(x^3),$$
whence $\;\tan x-\sin x\sim_0\dfrac{x^3}2$.
On the other hand, $\sqrt{1 + \tan(x)} + \sqrt{1 + \sin(x)}\xrightarrow[x\to 0]{}2$ , so we have
$$\frac{\tan(x) - \sin(x)}{x^3\bigl(\sqrt{1 + \tan(x)} + \sqrt{1 + \sin(x)}\bigr)}\sim_0 \frac{\frac12\, x^3}{x^3\cdot 2}=\frac14.$$
A: Hint: Picking from your second approach,
$$\tan x-\sin x=\frac{\sin x(1-\cos x)}{\cos x}=\frac{2\sin^2(\frac x2)\sin x}{\cos x}$$
Now use the fact $\lim_{t\to0}\frac{\sin t}t=1$.
The answer is NOT $0$.

Edit: Using the above substitution you get$$\lim_{x\to0}\frac{2}{\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\cdot\frac{\sin^2(\frac x2)\sin x}{x^3}\\=\lim_{x\to0}\frac{2}{\cos x(\sqrt{1+\tan x}+\sqrt{1+\sin x})}\cdot\lim_{x\to0}\frac{\sin^2(\frac x2)\sin x}{x^3}$$
The first limit evaluates to $1$. Write the second limit as$$\lim_{x\to0}\frac{\sin x}x\cdot\lim_{t\to0}\frac{\sin t}{2t}\cdot\lim_{t\to0}\frac{\sin t}{2t}$$where $t=x/2\to0$. This evaluates to $1/4$.

