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Thread: epsilon delta definition

  1. #1
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    epsilon delta definition

    prove that $\displaystyle \lim_{x\to0}\frac{sinx}{x}=1$ by ($\displaystyle \epsilon$, $\displaystyle \delta$) definition
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  2. #2
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    Are well known the trigonometric inequalities...

    $\displaystyle \sin x < x < \tan x$ , $\displaystyle 0< x < \frac{\pi}{2}$ (1)

    Now deviding each term of (1) by $\displaystyle \sin x$ You obtain...

    $\displaystyle \displaystyle 1 < \frac{x}{\sin x} < \frac{1}{\cos x} \implies \cos x< \frac{\sin x}{x} < 1 $ (2)

    ... and because the first and third term of (2) tends to $\displaystyle 1$ if $\displaystyle x \rightarrow 0+$ , then is...

    $\displaystyle \displaystyle \lim_{x \rightarrow 0+} \frac{\sin x}{x} = 1$ (3)

    With similar procedure You can demonstrate that the same hold for $\displaystyle x \rightarrow 0-$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    thank you, but i am trying to prove this limit using epsilon delta definition. the information that you gave me is very helpful if i was trying to prove this limit using the squeezing theorem for example.

    in delta epsilon definition, basically we are trying to write the expression $\displaystyle \left|\frac{sinx}{x}-1|\right |$ as $\displaystyle k\left|x \right|$ where k is a number.
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  4. #4
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    anybody?
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