<s>[INST] 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?
 [/INST] $1-\cos{x} = 2\sin^2{\frac{1}{2}x}$, 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 $1/2$.
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