how do you find the lim of ( cosec^2 x -2)/ ( cot x -1 ) as x tends to pi/4?
thanks!
Remember that $\displaystyle \sin^2{x} + \cos^2{x} = 1$
Dividing both sides by $\displaystyle \sin^2{x}$ gives us
$\displaystyle 1 + \cot^2{x} = \csc^2{x}$.
So $\displaystyle \frac{\csc^2{x} - 2}{\cot{x} - 1} = \frac{1 + \cot^2{x} - 2}{\cot{x} - 1}$
$\displaystyle = \frac{\cot^2{x} - 1}{\cot{x} - 1}$
$\displaystyle = \frac{(\cot{x} - 1)(\cot{x} + 1)}{\cot{x} - 1}$
$\displaystyle = \cot{x} + 1$.
So $\displaystyle \lim_{x \to \frac{\pi}{4}}\frac{\csc^2{x} - 2}{\cot{x} - 1} = \lim_{x \to \frac{\pi}{4}}(\cot{x} + 1)$
$\displaystyle = \cot{\frac{\pi}{4}} + 1$
$\displaystyle = 1 + 1$
$\displaystyle = 2$.