Limit of composite trigonometric function

Question

I am trying to find an easy way to compute the limit as $x \to 0$ of

$$f(x) = \frac{\sqrt{1+\tan(x)} - \sqrt{1+\sin(x)}}{x^3}$$

from first principles (i.e. without using l'Hôspital's rule).

I have gone as far as boiling down the problem to computing the limit as $x \to 0$ of

$$\frac{1 - \cos(x)}{x^2}$$

I thought about using the Small Angle Approximation for cosine, which indeed gives the right answer but doesn't seem to be a very formal.

Any hint?

Also, my working was fairly long so if you have a straightforward way to compute the limit of $f(x)$ I would love to hear it :)

Answer 1

$$\frac{1-\cos(x)}{x^2}$$ $$=\frac{2\sin^2(x/2)}{x^2}$$ $$\to ?$$

For $x$ small enough , we have:

$$\tan x>x>\sin x$$(using the geometric interpretation)

then

$$\cos(x)<\frac{\sin x}{x}<1$$

and since the function $\cos x$ is a continuous function $$\lim_{x\to 0}\cos x=cos 0=1$$

apply this to above, also recall squeeze theorem, we get what you want.

How to prove $\cos x$ is continunous, you may ask. $$|\cos x-\cos y|=2|\sin(\frac{x+y}{2})\sin(\frac{x-y}{2})|\le2|\sin(\frac{x-y}{2})|$$ now, we only have to prove $$\lim_{x\to 0}\sin x=0$$ for x small enough,

$$0<|\sin x|<|x|$$

then

$$0\le \lim_{x\to 0}|\sin x|\le \lim_{x\to 0}|x|=0$$

Answer 2

Try using double-angle formulae.

$1 - \cos x = 1 - \left ( 1- 2 \sin^2 \left ( \frac{\theta}{2} \right ) \right ) = 2 \sin^2 \left ( \frac{\theta}{2} \right )$