Proving Limits

  • Thread starter Anakin_k
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  • #1
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Hello everyone, I am in need of a little assistance. I have a homework assignment due soon that consists of 5 questions. Of which, I have done all but the 4th one. I did start on it but I'm not sure where to go from there. I also would like for someone to confirm my solutions for question 1 and 5, as I'm not too sure about them (I am confident with Q2 and Q3).

Here are the questions: http://i56.tinypic.com/30nb1xf.png

Question 1 solution: http://i52.tinypic.com/2hfht83.jpg
Question 5 solution: http://i56.tinypic.com/2jdhlzl.jpg
Question 4 beginning: http://i53.tinypic.com/15dvs6w.jpg

Thank you.
 

Answers and Replies

  • #2
I can attempt to help on Question 4:

Proof:
Let [itex]\epsilon > 0[/itex]. Since [itex]\lim_{x \rightarrow \infty} g(x) = l[/itex], there exists an [itex]N_0[/itex] such that if [itex]x > N_0[/itex], then [itex]|g(x) - l| < \epsilon/2[/itex] (*).

Similary, there exists an [itex]N_1[/itex] such that if [itex]x > N_1[/itex], then [itex]|f(x)-m|<\epsilon/(2|-a|)[/itex]. It follows from the last inequality that [itex]|-af(x)+am|<\epsilon/2[/itex] (**).

Now choose [itex]N=\max\{N_0, \, N_1\}[/itex]. So if [itex]x > N[/itex], inequalities (*) and (**) must hold. And adding (*) with (**), we get

[itex]|g(x)-l| + |-af(x)+am| < \epsilon/2 + \epsilon/2 = \epsilon[/itex].

But by the triangle inequality,

[itex]|(g(x)-l) +(-af(x)+am)| \leq |g(x)-l| + |-af(x)+am|[/itex].

Rearranging terms on the left-hand side of the above inequality, I think you get what you want: If [itex]x > N[/itex], then [itex]|(g(x)-af(x))-(l-am)| < \epsilon[/itex].

I hope this helps.

P.S. I forgot, this proof assumes [itex]a[/itex] does not equal zero. In the case it does equal zero, the problem reduces down to showing [itex]\lim_{x \rightarrow}(g(x)-0f(x)) = \lim_{x \rightarrow}(g(x)-0) = \lim_{x \rightarrow}g(x) = l - 0m = l[/itex], and this is already true since it's given.
 
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  • #3
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That was extremely thorough Mathstatnoob, thank you for taking the time to post the answer. I appreciate it very much. Through your steps, I've learned how to manipulate little pieces here and there to arrive at the needed answer. :)

If anyone could confirm my Q1 and Q5 solutions, I'd be thankful.
 
  • #4
hunt_mat
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Question 1 looks okay to me. for Q5, this is just the product lemma from the algebra of limits.
 
  • #5
berkeman
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I can attempt to help on Question 4:

Proof:
Let [itex]\epsilon > 0[/itex]. Since [itex]\lim_{x \rightarrow \infty} g(x) = l[/itex], there exists an [itex]N_0[/itex] such that if [itex]x > N_0[/itex], then [itex]|g(x) - l| < \epsilon/2[/itex] (*).

Similary, there exists an [itex]N_1[/itex] such that if [itex]x > N_1[/itex], then [itex]|f(x)-m|<\epsilon/(2|-a|)[/itex]. It follows from the last inequality that [itex]|-af(x)+am|<\epsilon/2[/itex] (**).

Now choose [itex]N=\max\{N_0, \, N_1\}[/itex]. So if [itex]x > N[/itex], inequalities (*) and (**) must hold. And adding (*) with (**), we get

[itex]|g(x)-l| + |-af(x)+am| < \epsilon/2 + \epsilon/2 = \epsilon[/itex].

But by the triangle inequality,

[itex]|(g(x)-l) +(-af(x)+am)| \leq |g(x)-l| + |-af(x)+am|[/itex].

Rearranging terms on the left-hand side of the above inequality, I think you get what you want: If [itex]x > N[/itex], then [itex]|(g(x)-af(x))-(l-am)| < \epsilon[/itex].

I hope this helps.

P.S. I forgot, this proof assumes [itex]a[/itex] does not equal zero. In the case it does equal zero, the problem reduces down to showing [itex]\lim_{x \rightarrow}(g(x)-0f(x)) = \lim_{x \rightarrow}(g(x)-0) = \lim_{x \rightarrow}g(x) = l - 0m = l[/itex], and this is already true since it's given.

Please re-read the Rules link at the top of the page. Posting solutions to homework questions is not permitted here. Instead, please provide hints, ask questions, find mistakes, etc.
 

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