1. ## limit question

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At first I was thinking of using the $\frac{sinx}{x}$ or the $\frac{cosx-1}{x}$, but this limit isn't going to 0. How should I go about solving this one?

2. May be you can find useful the following identity...

$\frac{5-5\cdot \tan x}{\sin x - \cos x}= - \frac{5}{\cos x}\cdot \frac{1-\tan x}{1-\tan x} = -\frac{5}{\cos x}$

... that produces a function that is 'well defined' in $x=\frac{\pi}{4}$ ...

Kind regards

$\chi$ $\sigma$

3. ahhh, I see now. Very useful indeed.

4. Originally Posted by hazecraze
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At first I was thinking of using the $\frac{sinx}{x}$ or the $\frac{cosx-1}{x}$, but this limit isn't going to 0. How should I go about solving this one?

Hint: $\frac{1-\frac{\sin x}{\cos x}}{\sin x-\cos x}=-\frac{1}{\cos x}\longrightarrow_{x\rightarrow\frac{\pi}{4}}\,-\sqrt{2}$

Tonio