why is the lim x->0 sin(x)/x = 1 and also why does this series diverges?

Question

this is the series
summation from n=1 to infinity of (n sin (1/n))

Answer 1

Part I (Solution without using L'Hopital's rule)

In a unit circle, draw a small angle x, x > 0

sinx = y

x = corresponding arc length

tanx = is the height of the right triangle of the base which is 1 unit.

So, you have

sinx ≤ x ≤ tanx

Divding both sides by sinx,

1 ≤ x/sinx ≤ 1/cosx

Flip all sides,

cosx ≤ sinx/x ≤ 1

Take limit of x->0, and by squeeze theorem,

lim{x->0} sinx/x = 1, since lim{x->0}cosx = 1



Part II

Since the general term n/sin(1/n) doesn't approach zero as n->∞, the series diverges.

------------

Work for part II:

lim{n->∞} n/sin(1/n)

= lim{x->0} 1/[sin(x)/x], where x = 1/n

= 1

Answer 2

You can prove it analytically if you know about l'Hôpital's Rule, but there's also a geometric argument that is fairly neat: take a sector (i.e. "pie-slice") of a circle with radius 1 and subtending an angle of x radians at the centre. The two radii are joined by a curved line, but you can also join them with a straight line. The length of the curved line is x, and of the straight line is sin x (prove it!), and as you shrink x the two lines become closer in length.



Then, because of that limit, the limit of xsin(1/x) as x approaches infinity also approaches 1, so the sum goes towards an infinite sum of 1s, which is infinite.

Answer 3

you can use L'Hopital's rule for the limit part of the question.



lim [x->0] (sin(x) / x) = lim [x->0] (cos(x) / 1) = cos(0) = 1



not sure about the second bit. not very good at the series analysis stuff. sorry!

Answer 4

sin(0) = 0. By theory, anything over anything = 1. correct? Except for zero/zero, which is a division by zero and therefore undefined. That's why we represent it as "lim x -> 0" which means as x tends to zero. The closer x is to zero, the more true this statement gets, except that x isn't zero!



as x gets closer to zero, since there's a division by x, this division becomes a division by a smaller and smaller number, whereas sin(x) cannot go beyond {-1:1}. If you're dividing something by a small number, answer becomes large. So you're answer is getting larger and larger thus "diverging".

Answer 5

No idea on the second one. The first one you use l'hopital's rule:

lim = f(x)

------ where the answer is indeterminate

g(x)

lim = f'(x)

------

g'(x)



So

lim = Sin(x)

x->0 ------

x

lim = Cos(x) = 1

x->0 ------

1