1. ## Limits questions

Some questions regarding limits:

1-$\displaystyle \lim(\sin(x)$$\displaystyle + x)/x \displaystyle x\rightarrow+\infty In this case, can I say that : 1-\displaystyle \lim(\sin(x) )\displaystyle /x$$\displaystyle = 0$ $\displaystyle + 1 = 1 ?$
$\displaystyle x\rightarrow+\infty$

I mean, I know that $\displaystyle sin (x)$ when x approaches infinite does not exist, but since it will have a maximum range of 1~-1 I can say it is aproximadely 0?
2-$\displaystyle \lim(\cos(x)$ )$\displaystyle /(\frac{\pi}{2} - x)$
$\displaystyle x\rightarrow\frac{\pi}{2}$

Here I dont have a single clue on how to do, seriously.

2. 1) is correct

you can show that sin x divided by x goes to zero as x goes to infinity
by the squeeze theorem

$\displaystyle {-1\over x}\le {\sin x\over x}\le {1\over x}$

2) can be done by L'Hopital's Rule. Have you seen that yet?
Otherwise you can let u=pi/2-x and shift this.

3. Try L'Hospitals Rule for 2.

4. Originally Posted by Capes
2-$\displaystyle \lim(\cos(x)$ )$\displaystyle /(\frac{\pi}{2} - x)$
$\displaystyle x\rightarrow\frac{\pi}{2}$
The most secure way uses the basic trigonometric identity $\displaystyle \cos x = \sin (\frac{\pi}{2} - x)$ that reports to the well known $\displaystyle \displaystyle \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$