Why does the limit of sin(dx/2)/(dx/2) equal 1?

[thereddevils]

In one of the examples in my book , it says that

$$\lim_{\delta x\rightarrow 0}\frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}=1$$

how can that be ?

 

[Mark44]

In one of the examples in my book , it says that

$$\lim_{\delta x\rightarrow 0}\frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}}=1$$

how can that be ?

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.

$$\lim_{x \rightarrow 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=1$$

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]

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.

$$\lim_{x \rightarrow 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}=1$$

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.

thanks Mark , but why must it be in radian mode ?

 

[Dick]

thanks Mark , but why must it be in radian mode ?

Because sin(x)~x for x small only works if x is radians. sin(1 degree) is nowhere near 1.