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?
Best Answer
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 $$\frac{\tan(x) - \sin(x)}{x^3\bigl(\sqrt{1 + \tan(x)} + \sqrt{1 + \sin(x)}\bigr)}\sim_0 \frac{\frac12\, x^3}{x^3\cdot 2}=\frac14.$$