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