Need Help with Homework: Question 1, 4 and 5

[Anakin_k]

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.

 

[mathstatnoob]

I can attempt to help on Question 4:

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

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

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

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

But by the triangle inequality,

$|(g(x)-l) +(-af(x)+am)| \leq |g(x)-l| + |-af(x)+am|$.

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

I hope this helps.

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

 

[Anakin_k]

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.

 

[hunt_mat]

Question 1 looks okay to me. for Q5, this is just the product lemma from the algebra of limits.

 

[berkeman]

I can attempt to help on Question 4:

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

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

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

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

But by the triangle inequality,

$|(g(x)-l) +(-af(x)+am)| \leq |g(x)-l| + |-af(x)+am|$.

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

I hope this helps.

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

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