Why does the `lim_(x->0) sin(x)/x = 1`
Why is `lim_(x->0)(sin(x))/x=1 ` ?
We will use the squeeze theorem. Consider a right triangle with one side of length 1, and the acute angle formed by this side and the hypotenuse with measure `theta ` .(See attachment.)
Draw a circle with radius 1 and center the vertex of the angle marked `theta ` .
(1) The area of the triangle is 1/2 base times height or `(1/2)(1)(tan(theta))=tan(theta)/2 `
(2) The area of the sector determined by the sides of the acute angle is ` 1/2 r^2 theta=theta/2 `
(3) If we drop a perpendicular from the point of intersection of the circular arc to the side of length 1, and form a triangle using the points of intersection of the arc and the sides of the original right triangle we find that the area of this smaller triangle is ` (1/2)(1)(sin(theta))=sin(theta)/2 `
Now from the geometry we have:
`tan(theta)/2 >= theta/2 >= sin(theta)/2 ` Multiply through by ` 2/sin(theta) ` to get:
`1/cos(theta)>= sin(theta)/theta >= 1 ` or
`cos(theta)<=sin(theta)/theta <= 1 `
Note that `cos(-theta)=cos(theta) ` and `sin(-theta)/(-theta)=sin(theta)/theta ` , so the inequality holds for all nonzero `theta in ` `(-pi/2,pi/2) ` .
Now `lim_(x->0)cos(theta)=lim_(x->0)1=1 ` , so by the squeeze theorem `lim_(x->0)sin(theta)/theta=1 `
Another way to show that
`lim_(x->0) sinx/x = 1`
is by using the L'Hospital rule. According to this rule,
`lim f(x)/g(x) = lim (f'(x))/(g'(x))` .
That is, the limit of the ratios of functions equals to the limit of the derivatives, if this limit exists.
In the case of sin(x) and x,
(sin(x))' = cos(x)
and (x)' = 1.
So, since cos(0) = 1,
`lim_(x->0) sinx/x = lim_(x->0) cosx/1 = 1`