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Math Help - limit question

  1. #1
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    limit question

    .

    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?
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    MHF Contributor chisigma's Avatar
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    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
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    ahhh, I see now. Very useful indeed.
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    Quote Originally Posted by hazecraze View Post
    .

    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
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