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At first I was thinking of using the $\displaystyle \frac{sinx}{x}$ or the $\displaystyle \frac{cosx-1}{x}$, but this limit isn't going to 0. How should I go about solving this one?
May be you can find useful the following identity...
$\displaystyle \frac{5-5\cdot \tan x}{\sin x - \cos x}= - \frac{5}{\cos x}\cdot \frac{1-\tan x}{1-\tan x} = -\frac{5}{\cos x}$
... that produces a function that is 'well defined' in $\displaystyle x=\frac{\pi}{4}$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$