The discussion revolves around the limit of the expression \(\lim_{x \rightarrow 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=1\), exploring the reasoning behind this limit in the context of trigonometric functions and their behavior as the variable approaches zero.
The discussion is ongoing, with participants exploring the implications of using radians versus degrees in the context of the limit. There is an acknowledgment of the approximation's validity for small angles in radians, but no consensus has been reached on all aspects of the limit's interpretation.
Participants note the challenge of typing Greek letters and the preference for simpler notation. There is also mention of using calculators to verify the behavior of the function as \(x\) approaches zero.
thereddevils said:In one of the examples in my book , it says that
[tex]\lim_{\delta x\rightarrow 0}\frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}=1[/tex]
how can that be ?
Mark44 said:Let's make the limit a little simpler by getting rid of the Greek letters. They don't really add anything and it's a pain to have to type in \delta all the time.
[tex]\lim_{x \rightarrow 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=1[/tex]
This can be and is because for values of x that are close to zero, sin(x) is approximately equal to x, making the ratio close to 1. You can convince yourself of the reasonableness of my claim by using your calculator (in radian mode) to calculate sin(x)/x for x = .1, .01, .001, and so on (also for negative x that is close to 0).
In this problem, if x is close to zero, then x/2 will be even closer to zero, so the ratio sin(x/2)/(x/2) will be even closer to 1 than would be the ratio of sin(x)/x.
Note that I am not proving anything here.
thereddevils said:thanks Mark , but why must it be in radian mode ?