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Delta-Epsilon Limit Proofs

Date: 09/05/98 at 02:43:11
From: Maylyn Angeles
Subject: Calculus

Thanks for helping me last time with the question you helped me figure 
out. Here are four more from my homework which I just couldn't figure:  

  22. For the limit as x approaches 2 of (4x+1)/(3x-4) = 4.5, 
      illustrate the definiton of a limit by finding the values of 
      delta that correspond to epsilon = 0.5 and epsilon = 0.1.

  30. Prove that the limit as x approaches 4 of (5-2x) = -3 using the  
      epsilon, delta definiton of a limit. Include a diagram.

  34. Prove that the limit as x goes to 2 of (x^2+x-6)/(x-2) using the 
      epsilon, delta definiton of a limit.

  42. Prove that the limit as x goes to -4 of (x^2-1) = 15 using the 
      epsilon, delta definition of a limit.

I hope this isn't too much, but it's these sorts of problems that I 
seem to be having the most trouble with. If you can help, thanks.


Date: 09/05/98 at 08:05:14
From: Doctor Jerry
Subject: Re: Calculus

Hi Maylyn,

I'll try to answer close to the question.

Question 22:
Write (4x+1)/(3x-4) - 4.5 = (4x+1)/(3x-4) - 9/2 and combine the 

   |(4x+1)/(3x-4) - 4.5| = (19/2)|x-2|/|3x-4|

When x is within 1 of 2, that is |x-2| < 1, the denominator |3x-4| is 
not smaller than 1. So:

      (19/2)|x-2|/|3x-4| < (19/2)|x-2| if delta is less than 1

So, to force (19/2)|x-2| to be less than 0.5: 

   |x-2| < 0.5*2/19.

So, we can take delta to be anything less than 1/19, say 1/20.

Question 30:
Since |-3-(5-2x)| = |-8+2x| = 4|x-2|, to force 4|x-2| < E, it is 
enough to take |x-2| < E/4.

Question 34:
Since we won't allow x to be 2, we can say that:

   (x^2+x-6)/(x-2) = x+3 

and so the limit would be 5 as x->2.

Since |5-(x+3)| = |x-2|, to make |x-2| < E, it is enough to take 
|x-2| < E.

Question 42:
Note that |15-(x^2-1)|=|16-x^2|=|4-x||4+x|.

Let's first control the annoying factor |4+x|. We do this by making a 
preliminary condition on delta. If delta is less than one, then if 
|x-4| < delta, we see that |x+4| is smaller than 5. So:

   |15-(x^2-1)| = |16-x^2| = |4-x||4+x| < 5|x-4|

To force 5|x-4| < E or |x-4| < E/5, it is enough to take delta = E/5 
and delta less than 1. So, the "formula" for delta is:  

   If E>0 is given, choose delta to be less than E/5 and less than 1.

- Doctor Jerry, The Math Forum
  Check out our web site! http://mathforum.org/dr.math/   
Associated Topics:
High School Analysis
High School Calculus

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