Limits of sin 1/x

craigmain · #1 $Old$ January 3rd, 2009, 09:39 AM

Hi,

I am reading spivaks' book and he states (p91) that abs(sin 1/x) <= 1 for all x, which is quite obviously true. How do you prove it though.
I have tried to reason it using the unit circle definition of sin but I cannot grasp why this is always true. Its obvious from the graph, but i am hoping there is a more formal reason.

Thanks
Regards
Craig.

Ziaris · #2 $Old$ January 3rd, 2009, 10:11 AM

Note that $\sin(f(x))$ always returns a value in $[-1,1]$ for a real function of a real variable $f(x).$ This is because no matter what number the function $f$ puts out, it will be some real number. You know that the normal graph of $\sin x$ is always in between -1 and 1, to the limit at negative and positive infinity. So no matter what real number $f$ produces, the value of $\sin(f(x))$ will still be in $[-1,1]$ (is there a counterexample to this?) Anyways, your function $1/x$ does behave in that fashion. Therefore the absolute value simply restricts the range of the function, such that $\left|\sin\left(\frac{1}{x}\right)\right|$ always returns a value in $[0,1].$

hatsoff · #3 $Old$ January 3rd, 2009, 10:20 AM

Quote:

Originally Posted by craigmain $View Post$

Hi,

I am reading spivaks' book and he states (p91) that abs(sin 1/x) <= 1 for all x, which is quite obviously true. How do you prove it though.
I have tried to reason it using the unit circle definition of sin but I cannot grasp why this is always true. Its obvious from the graph, but i am hoping there is a more formal reason.

Thanks
Regards
Craig.

The domain of the $g(x)=|\sin{x}|$ is $[0,1]$ . So, as long as the domain of the argument in the sin function is the same or more constricted as $\mathbb{R}$ , that function will itself have the same or more constricted domain.

craigmain · #4 $Old$ January 3rd, 2009, 11:51 PM

Thanks.
That's exactly what I needed to know. Its frustrating how stating the obvious is so darn difficult so often.