1. ## trigonometric limit

$\displaystyle \lim_{x \to 0}\frac{x-tan2x}{sin2x}$

2. Originally Posted by FLTR
$\displaystyle \lim_{x \to 0}\frac{x-tan2x}{sin2x}$
The function,
$\displaystyle f(x)=\frac{x-\tan 2x}{\sin 2x}$
Can be expressed equivalently as,
$\displaystyle f(x)=\frac{x}{\sin 2x}-\frac{\tan 2x}{\sin 2x}$
If we "cancel" to get,
$\displaystyle g(x)=\frac{x}{\sin 2x}-\sec 2x$
Then the functions, $\displaystyle f(x),g(x)$ will agree on some open interval containing $\displaystyle 0$ except possiblly at $\displaystyle x=0$ itself.
Thus, the limit exististence and value is equivalent to,
$\displaystyle \lim_{x\to 0} \frac{x}{\sin 2x} - \sec 2x$
Express as,
$\displaystyle \lim_{x\to 0} \frac{1}{2}\cdot \frac{2x}{\sin 2x} - \sec 2x$
We note that,
$\displaystyle \lim_{x\to 0} \frac{2x}{\sin 2x}=1$ by the limit composition rule and one of the famous limits.
Thus, since the functional limits exists so does its sum which is,
$\displaystyle \frac{1}{2} \lim_{x\to 0}\frac{2x}{\sin 2x} - \lim_{x\to 0}\sec 2x=\frac{1}{2}-1=-\frac{1}{2}$

3. Taylor works fast here too, sin(a) =~ tan(a) =~ a for a arround 0, so:

$\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{x - \tan 2x}}{{\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - 2x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{2x}} = - \frac{1}{2}$