I understand the epsilon-delta definition, but I can't work out problems with sines. Can someone help me with this exercises:

I have to solve:

1. sin(x) --> 1 for x --> pi/2

then abs(x-pi/2)<delta, but how do I work out abs(sin(x)-pi/2)<epsilon ??

2. (pi+2)sin(x)-2x-2 -->0 for x --> pi/2

how do I work out abs((pi+2)sin(x)-2x-2)<epsilon ??

thx for the help

Dear sung,

Proof of \(\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}{\sin{x}=1}\)

Recall the \(\displaystyle \epsilon-\delta\) definition. We have to show that, \(\displaystyle \forall~\epsilon>0~\exists~\delta>0~such~that,~0<\left|~x-\frac{\pi}{2}\right|<\delta\Rightarrow\mid\sin{x}-1\mid<\epsilon\)

Consider, \(\displaystyle \mid\sin{x}-1\mid\)

\(\displaystyle =\left|\sin{x}-\sin\frac{\pi}{2}\right|\)

\(\displaystyle =\left|~2\sin\frac{\left(x-\frac{\pi}{2}\right)}{2}\cos\frac{\left(x+\frac{\pi}{2}\right)}{2}\right|\)

\(\displaystyle =\left|~2~\right|\left|\sin\frac{\left(x-\frac{\pi}{2}\right)}{2}\right|\left|\cos\frac{\left(x+\frac{\pi}{2}\right)}{2}\right|\)

\(\displaystyle <\mid~2~\mid\left|\sin\frac{\left(x-\frac{\pi}{2}\right)}{2}\right|\) ; since \(\displaystyle \left|\cos\frac{\left(x+\frac{\pi}{2}\right)}{2}\right|>0\)

\(\displaystyle <\mid~2~\mid\left|\frac{\left(x-\frac{\pi}{2}\right)}{2}\right|\) ; since, \(\displaystyle \mid\sin{u}\mid<\mid{u}\mid\forall~u\in{R}~where~u~is~in~radians\)

Take any \(\displaystyle \epsilon\) such that,

\(\displaystyle \mid\sin{x}-1\mid<\left|{x-\frac{\pi}{2}}\right|<\epsilon\)

Therefore, \(\displaystyle \mid\sin{x}-1\mid<\epsilon~whenever~\left|~x-\frac{\pi}{2}\right|<\epsilon\)

Take \(\displaystyle \delta=\epsilon\) ;

\(\displaystyle \mid\sin{x}-1\mid<\epsilon~whenever~\left|{x-\frac{\pi}{2}}\right|<\delta\)

Hence according to the \(\displaystyle \epsilon-\delta\) definition,

\(\displaystyle \lim_{x\rightarrow\frac{\pi}{2}}{\sin{x}=1}\)

Hope this will help you.