Limit as x Approaches Infinity

[Jason76]

\(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

The page says the answer is \(\displaystyle -\dfrac{1}{4}\)

What is the first step in looking at this thing when \(\displaystyle x\) is not a number but infinity? If it's a number you just plug it in, and then see if the answer comes to \(\displaystyle 0/0\) (if not \(\displaystyle 0/0\), then you get your answer from the substitution). If it does, then you look for other ways to solve it.

 

[chalaman1403]

There are a couple of ways to see this, the most intuitive to me, is to neglect the lower order terms of each polynomial. Since x is approaching infinity, it is as huge as you can imagine and more, think that \(\displaystyle x^2 -4\) is virtually just \(\displaystyle x^2\) you can neglect the -4 because it is much smaller than \(\displaystyle x^2\), so you can consider the numerator to be just \(\displaystyle x^2\). Similar with the denominator, you can neglect the 2 and the x because as x is getting huge, they are nothing compared to the \(\displaystyle -4x^2\) so the denominator will be virtually just \(\displaystyle -4x^2\).
Then you just have \(\displaystyle \frac{x^2}{-4x^2}\). Do you see why is this equal to -1/4?
I hope that helped, regards
Damián Vallejo.

 

[stapel]

\(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

The page says the answer is \(\displaystyle -\dfrac{1}{4}\)

Think back to algebra, when you did horizontal asymptotes. What would the asymptote be for this?

Then think about the meaning of an horizonal asymptote, and compare this with "the limit as x gets very large".

What is the first step in looking at this thing when \(\displaystyle x\) is not a number but infinity?

You could apply one of the methods they gave you in class. One method commonly provided in standard textbooks is to "divide every term by the highest-index variable"; in this case, you'd divide everything by x^2. Then take the limit as x gets very large. What values will the fractions "take on"? What will you end up with? And so forth.

What methods did your textbook and instructor provide?

 

[JeffM]

\(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)

The page says the answer is \(\displaystyle -\dfrac{1}{4}\)

What is the first step in looking at this thing when \(\displaystyle x\) is not a number but infinity? If it's a number you just plug it in, and then see if the answer comes to \(\displaystyle 0/0\) (if not \(\displaystyle 0/0\), then you get your answer from the substitution). If it does, then you look for other ways to solve it.

Jason

I am not sure that I have understood your question. Are you asking what is the meaning of the limit of a function where the function's argument is "approaching infinity"? If so, you need to remember that this is just a shorthand way to say that the magnitude of x becomes an arbitrarily large but still finite number.

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}f(x) = a\ MEANS\ for\ any\ \epsilon > 0,\ \exists\ \delta > 0\ such\ that\ x > \delta \implies |f(x) - a| < \epsilon.\)

Formally, we are just dealing with large finite numbers.

 

[Jason76]

If you keep plugging values into x in the positive direction, then you have to see where y goes. y should approach \(\displaystyle -\dfrac{1}{4}\) according to this method.

 

[pka]

\(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}\)
The page says the answer is \(\displaystyle -\dfrac{1}{4}\)

@Jason76, do you agree that \(\displaystyle \displaystyle{\lim _{x \to \infty }}\dfrac{c}{x} = 0~?\)

If you do agree, then consider \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1-\frac{4}{x^2}}{\frac{2}{x^2}+\frac{1}{x}-4}\)
Then apply that limit.


Now if you don't agree, the you need to stop trying to understand infinite limits.

 

[Jason76]

@Jason76, do you agree that \(\displaystyle \displaystyle{\lim _{x \to \infty }}\dfrac{c}{x} = 0~?\)

If you do agree, then consider \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1-\frac{4}{x^2}}{\frac{2}{x^2}+\frac{1}{x}-4}\)
Then apply that limit.


Now if you don't agree, the you need to stop trying to understand infinite limits.

I know that in the case of \(\displaystyle c\) over \(\displaystyle x\) that \(\displaystyle y\) approaches \(\displaystyle 0\) as \(\displaystyle x\) approaches infinity. However, in this case, it approaches \(\displaystyle -\dfrac{1}{4}\), but how do we know that?

 

[chalaman1403]

I know that in the case of [FONT=MathJax_Math]c[/FONT] over [FONT=MathJax_Math]x[/FONT] that [FONT=MathJax_Math]y[/FONT] approaches [FONT=MathJax_Main]0[/FONT] as [FONT=MathJax_Math]x[/FONT] approaches infinity. However, in this case, it approaches -1/4[FONT=MathJax_Main][/FONT], but how do we know that?

Because -4/x^2 approaches 0, 2/x^2 approaches 0 and 1/x approaches 0, so all you have left is 1 in the numerator, and -4 in the denominator.

 

[Jason76]

@Jason76, do you agree that \(\displaystyle \displaystyle{\lim _{x \to \infty }}\dfrac{c}{x} = 0~?\)

If you do agree, then consider \(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1-\frac{4}{x^2}}{\frac{2}{x^2}+\frac{1}{x}-4}\)
Then apply that limit.


Now if you don't agree, the you need to stop trying to understand infinite limits.

How did the equation change (and why) from the way it originally looked.

\(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1-\frac{4}{x^2}}{\frac{2}{x^2}+\frac{1}{x}-4}\)

 

[chalaman1403]

How did the equation change (and why) from the way it originally looked.

\(\displaystyle \dfrac{x^{2} - 4}{2 + x - 4x^{2}}=\dfrac{1-\frac{4}{x^2}}{\frac{2}{x^2}+\frac{1}{x}-4}\)

Divide both numerator and denominator by x^2, remember that you can multiply or divide both numerator and denominator by a same thing, the change was so that we get all those terms to go to 0. What I told you in my first answer (neglecting the lower order terms) is what I find to be more intuitive though.

 

[JeffM]

If you keep plugging values into x in the positive direction, then you have to see where y goes. y should approach \(\displaystyle -\dfrac{1}{4}\) according to this method.

Which is exactly what your book says, right?

[FONT=MathJax_Main]lim[/FONT] as [FONT=MathJax_Math]x[/FONT] approaches infinity \(\displaystyle \dfrac{x^2 - 4}{2 + x - 4x^2}.\)

The page says the answer is \(\displaystyle -\dfrac{1}{4}\)

\(\displaystyle f(10,000) = \dfrac{10,000^2 - 4}{2 + 10,000 - 4 * 10,000^2} = \dfrac{100,000,000 - 4}{10,002 - 4 * 100,000,000} = \dfrac{99,999,996}{-399,989,998} = -0.250006 \approx - \dfrac{1}{4}.\)

If this is not close enough to -1/4 to satisfy you, pick a bigger number, say x = 1,000,000. The delta-epsilon method is just a way to make this experimental approach rigorous.

What chalaman and stapel have told you is the simplest and quickest method to deal with the limit of a rational function as its argument approaches infinity, but I am not sure that the method is more intuitive until you understand the general reasoning behind it.

Suppose f(x) is the ratio of two polynomials. That is:

\(\displaystyle f(x) = \dfrac{\displaystyle \sum_{i = 0}^ma_ix^i}{\displaystyle \sum_{i = 0}^nb_ix^i},\ a_m \ne 0,\ b_m \ne 0,\ m,\ n \in \mathbb I^+.\)

Now there are three possible cases. m = n. m > n. Or m < n.

\(\displaystyle m = n \implies f(x) = f(x) * 1 = f(x) * \dfrac{\frac{1}{x^n}}{\frac{1}{x^n}} = f(x) * \dfrac{\frac{1}{x^m}}{\frac{1}{x^n}} = \dfrac{\displaystyle a_m + \sum_{i = 0}^{m - 1}a_ix^{(i - m)}}{\displaystyle + b_m + \sum_{i=0}^{m-1}b_ix^{(i - m)}} \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}f(x) = \dfrac{a_m}{b_m}\) because the limit of each element of the two sums, and therefore the limit of each sum, approaches zero.

Using similar logic, you should be able to work out that:

\(\displaystyle m > n \implies \displaystyle \lim_{x \rightarrow \infty}f(x) = \infty.\)

\(\displaystyle m < n \implies \displaystyle \lim_{x \rightarrow \infty}f(x) = 0.\)

This takes us back to stapel's and chalaman's original posts, and the fundamental logic behind it was given in pka's original post.

 

[Jason76]

Perhaps a strategy (for these infinity problems) would be to just pick a few huge numbers and plug into x. For instance, one time you could pick 10,000, then another time 100,000. It should always come out near the target answer. Is this right? As far as the manipulation done to the problem (the way some people tried to solve it on here), I don't really understand it.

For instance, if we plugged in \(\displaystyle 10,000\), and later \(\displaystyle 100,000\) into \(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{1}{x}\), then we should get something near \(\displaystyle 0\) both times, right?

 

[Deleted member 4993]

The strategy should be what stapel and other instructors had instructed you for several posts:

"divide every term by the highest-index variable"; in this case, you'd divide everything by x^2. Then take the limit as x gets very large. What values will the fractions "take on"?

Then apply "your" strategy to see whether the answer is correct.

 

[JeffM]

Perhaps a strategy (for these infinity problems) would be to just pick a few huge numbers and plug into x. For instance, one time you could pick 10,000, then another time 100,000. It should always come out near the target answer. Is this right? As far as the manipulation done to the problem (the way some people tried to solve it on here), I don't really understand it.

For instance, if we plugged in \(\displaystyle 10,000\), and later \(\displaystyle 100,000\) into \(\displaystyle \lim\) as \(\displaystyle x\) approaches infinity \(\displaystyle \dfrac{1}{x}\), then we should get something near \(\displaystyle 0\) both times, right? Correct but

One way to do these problems is, as you say, to try various huge magnitudes and see what you get. There are several issues with this approach. One is that it is not clear what is "huge." Hugeness is relative to the specific problem. Another issue is that your answer is approximate. In the example I gave in my previous post, I calculated -0.25006, which is not the exact answer. A third, closely related issue, is that a positive proof can never be given by example. The big advantage of this method is that it can be used to check the results of a more rigorous method.

 

[Jason76]

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfdirectory/LimitInfinity.html#PROBLEM - Problems

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfsol1directory/LimInfSol1.html - Solutions

Here are some problems about "x going toward infinity".

I see the strategy behind what their doing, and could copy it, but I don't know why they are doing it.

But we can see some patterns:

• Negative or positive infinity divided by "something without infinity" is 0
• "Negative or postive Infinity - negative or positive infinity" indeterminate 0/0 (but you might be able to get a different answer if factored)
• "negative or positive Infinity + negative or positive infinity" is infinity
• "negative or positive infinity * negative or positive infinity" is infinity

Could be some mistakes here, correct me.

 

[JeffM]

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfdirectory/LimitInfinity.html#PROBLEM - Problems

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/liminfsol1directory/LimInfSol1.html - Solutions

Here are some problems about "x going toward infinity".

I see the strategy behind what their doing, and could copy it, but I don't know why they are doing it.

Jason

I am sorry, but please take a specific problem and explain what you do not understand about the given solution. (The one I looked at was highly informal, but quite intuitive.)

 

[Jason76]

Jason

I am sorry, but please take a specific problem and explain what you do not understand about the given solution. (The one I looked at was highly informal, but quite intuitive.)

I brought up the links for general info, not to solve a specific problem. But we can see some generalizations from looking at the problems (on the link).