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Proving Limits at Infinity Using the Formal Definition

Date: 04/11/2004 at 09:20:40
From: David
Subject: Calculus - Formal Definition of Limits as x -> infinity.

We are required to find a limit (informally), then prove that our 
answer is correct by using the formal definition of limits, which 
is "Limit of f(x) as x -> infinity is equal to L, iff we can find M (a
function of epsilon) such that x > M forces |f(x) - L| < epsilon."

Specific examples:

a) f(x) = (sin 3x)/(x^2 + 4)
b) f(x) = (x + 3)/(x^2 - 3)

I do get the idea of epsilon and M, but when it comes to using them to
prove, I really don't know where to start.  How can I find M (a
function of epsilon), and if I do find M, which x do I replace in to

|(sin 3x)/(x^2 + 4) - 0| < epsilon so, find M by equating to get
|(sin 3x)/(x^2 + 4)| = epsilon.  From here I am not able to proceed.

Date: 04/11/2004 at 11:24:04
From: Doctor Fenton
Subject: Re: Calculus - Formal Definition of Limits as x -> infinity.

Hi David,

Thanks for writing to Dr. Math.  With limits, "equating" the two sides
usually doesn't help much, because you usually get an equation which
you can't solve analytically.  It is usually better to use estimation 
and work with inequalities.

For your first limit

   lim  ------- 
  x->oo x^2 + 4

it appears that you have correctly evaluated the limit, since for very 
large x, the numerator is a number between -1 and +1, while the 
denominator is an extremely large number, so the quotient will get 
smaller and smaller as the denominator gets larger and larger.

Formally, you have to show that given a positive epsilon (E), you can 
find an M such that if x > M, then

   | sin(3x)  |
   | -------- | < E
   | x^2 + 4  |

Because the left side is a very complicated expression, let's look for 
a simpler expression which is always larger than the complicated left 

If we increase the magnitude of the numerator of a fraction, the 
magnitude of the fraction increases, and if we decrease the magnitude
of the denominator of the fraction, again the magnitude of the 
fraction increases.

We know that |sin(3x)| <= 1, and |x^2 + 4| > x^2, so 

   | sin(3x) |    1
   | ------- | < ---
   | x^2 + 4 |   x^2

So, to make the left side < E, it is sufficient to make the right 
side, which is much simpler to analyze, less than E.

Can you now see how to choose M so that if x > M, then 1/x^2 < E?

The second problem will be similar, but you need to find something
larger than x + 3, so you might use

   x + 3 < 2x , for x > 3  (so that you will need to make sure M >= 3)

and you need something smaller than x^2 - 3, so you could use

             x^2                    x^2
   x^2 - 3 > ---  , as long as 3 < --- or x > sqrt(6)
              2                      2

If x > 3, to make the first inequality work, then x > sqrt(6), so this 
second inequality will also be true.

If you have any questions or need more help, please write back and 
show me what you have been able to do, and I will try to offer further

- Doctor Fenton, The Math Forum 

Date: 04/11/2004 at 11:40:21
From: David
Subject: Calculus - Formal Definition of Limits as x -> infinity.

Thanks for your reply.  I have done the examples according to your
suggestions, and have completed the set.  Here are my answers to the
second example; could you please confirm that I am on the right track
with your suggestions?  Thank you.

 |x + 3|         |4x|
---------  <  ----------  < E (for x > 3, and M >= 3 for x > 3)
|x^2 - 3|        |x^2|

                  ---     < E

thus:              4
                  ---     = M(E)

and if,                 x > M

                        1   1
                        - < -
                        x   M

                        4   4
                        - < -
                        x   M

which implies that | f(x) - L | < E, for x > M.

Date: 04/11/2004 at 12:05:56
From: Doctor Fenton
Subject: Re: Calculus - Formal Definition of Limits as x -> infinity.

Hi David,

That looks good.  Usually, to indicate that you want M >= 3 (to make
the estimating inequalities valid when x > M), and also M >= 4/E 
(to make the simplified expression less than E), you write 
M = max(3,4/E).  Then, if x > M,

the inequality

   |x + 3|         |4x|        4
  ---------  <  ----------  = ---
  |x^2 - 3|        |x^2|       x

is true, and since x >= 4/E, then 4/x < E, so

   |x + 3|         
  ---------  < E
  |x^2 - 3|      

- Doctor Fenton, The Math Forum 

Date: 04/11/2004 at 12:08:07
From: David
Subject: Thank you (Calculus - Formal Definition of Limits as x ->

Thanks very much for your reply. It was very helpful, and now I feel 
much better!

Thanks again :D
Associated Topics:
College Analysis
College Calculus
High School Analysis
High School Calculus

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