Expert: Paul Klarreich - 7/11/2007

Question
how do you prove(using the formal definition of a limit)  limit of sinx, as x approaches a, equals sina?

Answer
Questioner:   Julia
Category:  Calculus
 
Question:  how do you prove(using the formal definition of a limit)  limit of sinx, as x approaches a, equals sina?
..........................................
Hi, Julia,

These proofs always involve some tricky reasoning about inequalities, so I have two approaches here.  I hope you can follow at least one of them.

I am going to assume that you already know this limit:
      sin x
lim   ------ = 1
x->0     x

and this 'sum-to-product' trigonometric identity:

(cf:  http://www.sosmath.com/trig/Trig5/trig5/trig5.html )

sin u - sin v =  2 cos((u+v)/2) sin ((u-v)/2)

Lim  sin x = sin a means
x->a

Lim  sin x - sin a = 0
x->a

Given  epsilon, there exists a delta, [sorry, can't make a epsilon and delta on this computer, so 'e' and 'd' will have to do.]  such that

| sin x - sin a | < e   whenever  | x - a | < d

| sin x - sin a | = | 2 cos((x+a)/2) sin ((x-a)/2) |

| sin x - sin a | = 2 | cos((x+a)/2) sin ((x-a)/2) |

Now do some inequality work:

| sin x - sin a | <= 2 | sin ((x-a)/2) |

because  | cos (...) | <= 1

Now  x-a is our delta.  And since:
    sin t
lim  ------ = 1
t->0   t
     
that means for small values of t (how small?  How about less than delta?)  sin t = t.  That means for small values of  x-a:

sin((x-a)/2) is approximately (x-a)/2 = d/2.  Certainly we can find a 'd' such that sin((x-a)/2) < d, since it is near  d/2.  So our inequality is:

| sin x - sin a | <= 2 | sin ((x-a)/2) | <= 2d

OK, then.  All we have to do is make sure  d = e/2 and we have:

| sin x - sin a | <= e, whenever  |x-a| < d = e/2

.................ANOTHER APPROACH...............
Suppose we try this:

lim   sin x = sin a  means, if we let  x=a+h:
x->a

lim   sin(a+h) = sin a,  which means:
h->0

lim   sin(a+h) - sin a = 0
h->0

So we have to prove:

| sin(a+h) - sin a | < e, whenever  | h | < d

| sin a cos h + cos a sin h - sin a | =

| sin a (cos h - 1) + cos a sin h | =

| sin a (cos^2(h) - 1)/(cos h + 1) + cos a sin h |

The triangle inequality says that  |a+b| <= |a|+|b|:

| - sin a sin^2(h)/(cos h + 1) + cos a sin h | <=
| sin a sin^2(h)/(cos h + 1)| + | cos a sin h | <=

Now some inequality work:

If h is small, then

(a) cos h > 0, so

| sin a sin^2(h)/(cos h + 1)| <= | sin a sin^2(h) |

(b) sin x is approximately h, so is certainly  <= 2h

| sin a sin^2(h) | <= 4h^2 | sin a |

and  | sin a | <= 1, so

4h^2 | sin a | <= 4h^2

That's our first term.  The second term is:

| cos a sin h |

The same kind of reasoning gets us:

| cos a sin h | <= | sin h | <= 2h

So we finally have:

| sin(a+h) - sin a | <= 4h^2 + 2h = 2h(2h + 1)

and we can assume that  h < 1,

| sin(a+h) - sin a | <= 2h(2 + 1) = 6h

Ok, then.  Take that as our delta = 6h and we have our proof.