Limits: Division by 0, Making denominator = y

  • #1

Homework Statement



So we have just been learning limits which I understand, however there was one example which completely through me. It was regarding division by 0.

We had just been learning the known limits such as lim x->infinity (1/x) = 0 which I understood, and how anything similar such as 1/3x is 0 because its the same as 1/3 * 1/x which is basically multiplying by 0.


The problem was lim x->4+ x/(x-4), now the limit is +infinity, but I just thought it was this because 4/0 can relate to the known value of 1/x such as 4*1/0.
However the when the lecturer was explaining the proof, because the limit is for x->4+ we cant relate it to the known limits.

And the way she proved it was by letting y = x-4 (the denominator in the example) and then writing a new limit problem which was lim y->0+ (y+4)/y . The answer is still +infinity but this is the correct "proof" / logic, but I have a hard time understanding this concept.

Usually if I get stuck when I get home I just look up from other books and the net and I will find a good explanation of why, but in this case I dont even know what Im looking for, I know its to do with limits but I cant find anything similar anyway.

So if anyone could shed some light on this or perhaps this has a certain name for it...for example "solving by factorising" then I would love to know so I can start looking for the right thing.

I hope this makes sense, its the first time trying to post for help on this kind of thing.

Thanks
 
Last edited:

Answers and Replies

  • #2
AH, while looking for other posts based on limits so I knew which section to put my post in, I didnt realise that I was posting in the homework section.

As its not directly for homework or coursework, just for understanding the concept taught I guess its in the wrong section? If so, could it be moved to the appropriate section please.

Sorry about that, thanks
 
  • #3
34,893
6,633

Homework Statement



So we have just been learning limits which I understand, however there was one example which completely through me. It was regarding division by 0.

We had just been learning the known limits such as lim x->0+ (1/x) = +infinity which I understood, and how anything similar such as 1/3x is 0 because its the same as 1/3 * 1/x which is basically multiplying by 0.
How do you see this as multiplying by 0?
[tex]\lim_{x \to 0^+} \frac{1}{3x}= \lim_{x \to 0^+} \frac{1}{3} \frac{1}{x}[/tex]
[tex]= \frac{1}{3}\lim_{x \to 0^+} \frac{1}{x}[/tex]
You can split up limits like this as long as the separate limits exist. Here the limit of 1/x doesn't actually "exist" in the sense of being a specific finite number, but we still say that this limit is infinity.
The problem was lim x->4+ (x/x-4), now the limit is +infinity, but I just thought it was this because 4/0 can relate to the known value of 1/x such as 4*1/0.
4/0 is undefined. Also, you should get in the habit of putting parentheses in the right places. Knowledgeable people would interpret (x/x-4) as (x/x) - 4, which is not what you meant. Write this expression as x/(x - 4) if you want to be understood.
However the when the lecturer was explaining the proof, because the limit is for x->4+ we cant relate it to the known limits.

And the way she proved it was by letting y = x-4 (the denominator in the example) and then writing a new limit problem which was lim y->0+ (y+4/y) . The answer is still +infinity but this is the correct "proof" / logic, but I have a hard time understanding this concept.
All that happened was to change the variable used in the limit. If we make the substitution y = x - 4, we are in essence translating the graph of the limit function four units to the right. Instead of looking at the limit as x approaches 4+, we are now looking at the new limit as y approaches 0+. The limit value will be the same.

As already noted, use parentheses. (y+4/y) would be correctly interpreted as y + (4/y). What I think you mean is (y + 4)/y. Whenever a numerator or denominator contains more than one term, put parentheses around it.
Usually if I get stuck when I get home I just look up from other books and the net and I will find a good explanation of why, but in this case I dont even know what Im looking for, I know its to do with limits but I cant find anything similar anyway.

So if anyone could shed some light on this or perhaps this has a certain name for it...for example "solving by factorising" then I would love to know so I can start looking for the right thing.

I hope this makes sense, its the first time trying to post for help on this kind of thing.

Thanks
 
  • #4
525
16
In the x/(x-4) case, look at what's happening to x. x is approaching 4 from the right (or positive side), so values of x could be something like: 4.1, 4.01, 4.001, 4.0001, 4.00001, etc. Now when you're taking the limit, you're trying to figure out what happens to x/(x-4) as x gets arbitrarily closer and closer to four. Take those values I listed and plug them in and you'll get 4.1/0.1, 4.01/0.01, 4.001/0.001, etc. Note that the denominator is positive and getting closer and closer to zero, while the numerator is just getting closer to a constant: four, so you know that it approaches infinity. You know that it approaches positive infinity specifically by looking at the signs. Even though it approaches zero, it approaches zero from the positive side, so the denominator is always positive. The numerator is also positive, so you know that the expression as a whole is positive.
 
  • #5
Thanks for the reply Mark,

I see what you mean by the parentheses, however I was just using them to hold the actual equation to separate it from the text, not actually using them in a mathematical way. But I take it on board.

Ah sorry about the multiplication by zero, thats a bit of a typo on my part, what it meant to say was.

lim x->infinity for 1/x = 0. So 1/(3x) as x->infinity is still 0...as its like 1/3 * 1/x thus making it like multiplying by zero. Bit of a mix up on my part, ill see if I can change it in the original post.

In regards to your explanation of the problem, I still cant see the logic behind it. You have explained it for that example but I dont understand the reasoning and method of it so I would know how to implement it in any other problems if I would need to. Also I dont understand the reason of needing to prove it that way....

Also, does this method have any formal name?
 
  • #6
hunt_mat
Homework Helper
1,741
25
Look up the algebra of limits.
 
  • #7
34,893
6,633
Thanks for the reply Mark,

I see what you mean by the parentheses, however I was just using them to hold the actual equation to separate it from the text, not actually using them in a mathematical way. But I take it on board.

Ah sorry about the multiplication by zero, thats a bit of a typo on my part, what it meant to say was.

lim x->infinity for 1/x = 0. So 1/(3x) as x->infinity is still 0...as its like 1/3 * 1/x thus making it like multiplying by zero. Bit of a mix up on my part, ill see if I can change it in the original post.

In regards to your explanation of the problem, I still cant see the logic behind it. You have explained it for that example but I dont understand the reasoning and method of it so I would know how to implement it in any other problems if I would need to. Also I dont understand the reason of needing to prove it that way....

Also, does this method have any formal name?
I don't think so.

You might be making more out of this than it's really worth. All that's going on with the change of variables is converting a given problem into one whose limit is known. The graph of y = f(x) = x/(x - 4) has a vertical asymptote of x = 4. The graph of y = g(u) =(u + 4)/u has a vertical asymptote of u = 0. The only difference between these two graphs is a horizontal translation by four units.

[tex]\lim_{x \to 4^+}\frac{x}{x - 4} = \lim_{u \to 0^+}\frac{u + 4}{u}=\infty[/tex]
 

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