# epsilon delta definition

• Oct 15th 2010, 03:07 AM
eq123
epsilon delta definition
prove that $\lim_{x\to0}\frac{sinx}{x}=1$ by ( $\epsilon$, $\delta$) definition
• Oct 15th 2010, 03:39 AM
chisigma
Are well known the trigonometric inequalities...

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$
• Oct 15th 2010, 07:46 AM
eq123
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 $\left|\frac{sinx}{x}-1|\right |$ as $k\left|x \right|$ where k is a number.
• Oct 16th 2010, 08:26 AM
eq123
anybody?