# eps-delta for sines

#### sung

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

#### Sudharaka

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,

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