Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ $$\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$$
The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\sec(x) = \frac{1}{\cos(x)}$ we have
$$\frac{\frac{1}{\cos(x)}-1}{x^2\frac{1}{\cos(x)}}$$
And that is
$$\frac{\frac{1-\cos(x)}{\cos(x)}}{\frac{x^2}{\cos(x)}}$$
Then
$$\frac{(1-\cos(x))\cdot\cos(x)}{\cos(x) \cdot x^2}$$
And
$$\frac{(1-\cos(x))}{x^2}$$
Alright. It is my understanding that there is a formula or something that states $\frac{1-\cos(x)}{x} = 0$, so I can apply it here because we actually have
$$\frac{\color{green}{(1-\cos(x))}}{\color{green}{x}\cdot x}$$
Leaving us with
$$\color{green}0\cdot\frac{1}{x}$$
But if we evaluate, we get
$$0\cdot\frac{1}{0}$$
Which I guess is indeterminate because of $\frac{1}{0}$.
Also, the correct answer should be $\frac{1}{2}$.
What was wrong with my procedure?
 A: $1-\cos{x} = 2\sin^2{\frac{1}{2}x}$, so
$$ \frac{1-\cos{x}}{x^2} = \frac{1}{2}\left(\frac{\sin{\frac{1}{2}x}}{x/2}\right)^2, $$
and you probably know that
$$ \lim_{y \to 0} \frac{\sin{y}}{y} = 1, $$
so the limit is $1/2$.
A: Note that you get an undefined form :
$$
\lim_{x\to 0} \frac{1-\cos{(x)}}{x^2}=\frac{0}{0}
$$
So you can use l'Hopital rule to get:
$$
\lim_{x\to 0} \frac{1-\cos{(x)}}{x^2}=\lim_{x\to 0} \frac{\sin{(x)}}{2x}=\frac{1}{2}
$$
Because as you probably know :
$$
\lim_{x\to 0} \frac{\sin{(x)}}{x}={(\frac{d}{dx})}_{x=0} {\sin{(x)}}=cos{(0)}=1
$$
You can also use series expansion of $\cos{x}$ near $0$:
$$
\cos{(x)}=1-\frac{x^2}{2}+o(x^4)
$$
To get:
$$
\lim_{x\to 0}\frac{1-\cos{(x)}}{x^2}=\lim_{x\to 0}\frac{1-(1-\frac{x^2}{2}+o(x^4)))}{x^2}\\
=\lim_{x\to 0} \frac{\frac{x^2}{2}}{x^2}=\frac{1}{2}
$$
A: $$\cos(x)=1-\frac{x^2}{2}+o(x^2)$$
Then
$$\frac{1-\cos x}{x^2}=\frac{1}{2}+\varepsilon(x)$$
where $\varepsilon(x)\to 0$ when $x\to 0$.
A: $$\frac{1-\cos x}{x^2} = \frac{1}{1+\cos x}\frac{1-\cos^2 x}{x^2} = \frac{1}{1+\cos x}\frac{\sin^2 x}{x^2}  \to \frac{1}{2}\cdot 1^2 = \frac{1}{2}.$$
