prove that $\displaystyle \lim_{x\to0}\frac{sinx}{x}=1$ by ($\displaystyle \epsilon$, $\displaystyle \delta$) definition

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- Oct 15th 2010, 03:07 AMeq123epsilon delta definition
prove that $\displaystyle \lim_{x\to0}\frac{sinx}{x}=1$ by ($\displaystyle \epsilon$, $\displaystyle \delta$) definition

- Oct 15th 2010, 03:39 AMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 15th 2010, 07:46 AMeq123
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. - Oct 16th 2010, 08:26 AMeq123
anybody?