Delta-Epsilon Limit ProofsDate: 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. From, Maylyn 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 fractions: |(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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/