Limits

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alearon

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lim x->infinity
root [(9x^2 +2)/(4x^2 +3)]

I have no idea how to work out this limit. I still don't really get the concepts of limits. So if both the denominator and numerator = 0, it can have a limit? how does that work? And if the denominator only = 0, no real limit. What happens if numerator = 0 and denominator doesnt?

My friend worked it out by splitting it into two, then going:
9x^2 + 2 => 9 +2/(x^2) and putting the infinity into the x, making the value so small that you could remove it. Those two statements aren't equal though so I'm not sure how that works either... :(
 
Hello, alearon!

\(\displaystyle \lim_{x\to\infty}\sqrt{\frac{9x^2+2}{4x^2+3}}\)
Click to expand...

\(\displaystyle \text{Divide numerator and denominator by }x^2\!:\)

. . \(\displaystyle \lim_{x\to\infty}\sqrt{\frac{\frac{9x^2}{x^2} + \frac{2}{x^2}}{\frac{4x^2}{x^2} + \frac{3}{x^2}}} \;=\; \lim_{x\to\infty}\sqrt{\frac{9 + \frac{2}{x^2}}{4 + \frac{3}{x^2}}} \;=\;\sqrt{\frac{9+0}{4+0}} \;=\;\sqrt{\frac{9}{4}} \;=\;\frac{3}{2}\)

 


I note that soroban did not respond to any of the specific questions.

The suggestion to divide by x^2 is good, but made moot by providing camera-ready copy.

It's a shame the poster never got a chance to try.

 
alearon said:
lim x->infinity
root [(9x^2 +2)/(4x^2 +3)]

I have no idea how to work out this limit. I still don't really get the concepts of limits. So if both the denominator and numerator = 0, it can have a limit? how does that work? And if the denominator only = 0, no real limit. What happens if numerator = 0 and denominator doesnt?

My friend worked it out by splitting it into two, then going:
9x^2 + 2 => 9 +2/(x^2) and putting the infinity into the x, making the value so small that you could remove it. Those two statements aren't equal though so I'm not sure how that works either... :(
Click to expand...

The INTUITIVE idea about the limit at a finite number is pretty simple: if F(x) keeps getting closer to a as x gets closer to b then a is the limit of F(x) at b. The tricky conceptual part about limits is that F(b) itself is irrelevant. It's what happens very close to b that is relevant.

The INTUITIVE notion of a limit at infinity is only a little bit harder. If there exists a number c such that F(x) gets closer to a as
(x - c) gets bigger and bigger, then the limit of F(x) at infinity is a. Again what happens at infinity is irrelevant. It is what happens when x >= c.

Now let's answer your more specific CONCEPTUAL questions. Let's suppose F(x) = [G(x) / H(x)]. Your question above is: if G(b) = 0 and H(b) = 0, can F(x) have a limit at b? The answer is maybe yes or maybe no. Remember what happens at b is irrelevant. What matters is what is happening close to b. Your second question is: if G(b) = 0 and H(b) is not equal to 0, can F(x) have a limit at b? Again, it depends on what happens very close to b. If instead of limits at a number b, we are talking about limits at infinity, the same thing happens. What matters is what happens to G(x) and H(x) when x is very very large. Does this help?

I am going to attack your COMPUTATIONAL problem in a less elegant, but hopefully more comprehensible, way than soroban's.

A different way to express F(x) is to do the division called for by using the remainder method you learned back in 3rd or fourth grade.
In other words, F(x) = [(9x[sup:1ld9kmev]2[/sup:1ld9kmev] + 2) / (4x[sup:1ld9kmev]2[/sup:1ld9kmev] + 3)][sup:1ld9kmev](1/2)[/sup:1ld9kmev] = {(9/4) - [19 / 4(4x[sup:1ld9kmev]2[/sup:1ld9kmev] + 3]}[sup:1ld9kmev](1/2)[/sup:1ld9kmev].
Now the advantage of this expression is that it has two parts, one being (9/4) and other being [19 / (16x[sup:1ld9kmev]2[/sup:1ld9kmev] + 12)].
No matter how big x gets, it has no effect on (9/4). (9/4) = (9/4) til the cows come home. What about 19 / (16x[sup:1ld9kmev]2[/sup:1ld9kmev] + 12). Well as x gets big, that fraction gets very close to zero. Suppose x = 1000. Then x[sup:1ld9kmev]2[/sup:1ld9kmev] = 1,000,000 and 16x[sup:1ld9kmev]2[/sup:1ld9kmev] + 12 = 16,000,012. And 19 / 16,000,012 is pretty close to zero, right? Well what if x = 1,000,000. In that case, 16x[sup:1ld9kmev]2[/sup:1ld9kmev] + 12 = 16,000,000,000,012. Divide THAT into 19, and the result is tiny and is even closer to zero, isn't it. So the bigger x gets, the closer F(x) gets to (9/4)[sup:1ld9kmev](1/2)[/sup:1ld9kmev] = (3/2).

Does this make sense?
 
Thank you all for helping to explain, I definitely get the process for that particular question now.

@JeffM : I think I understand what you mean in regards to the intuitive notion of a limit at infinity - is c any number that could act as a 'grounder' as such in regards to infinity? So as x -> infinity, and the graphs limit is 1, c could be any number - 10, 100, 1000.. Just something to test how close the graph gets to 1?

And for the last part about the remainder method.. I have no idea how that works. I get how this would be easier to see the part where infinity makes the number too small to be significant however. What is the remainder method? I googled it, but it came up with some stuff that definitely wasn't it.

Thankyou again!
 
alearon said:
I have no idea how that works.
Click to expand...

For real, you simply must stop saying this and thinking this. Sometimes, mathematics will be a little like learning another language. It will take a little effort to think a little differently. Think about it. Let it roll around in your brain. Try some things. Draw some pictures. Get your hands dirty and see what you can produce. Don't EVER "have no idea" and just quit. It is likely to take considerable effort.

Note: The fact that you are having this discussion is very encouraging. Now, quit the personal sabotage. :)
 
alearon said:
Thank you all for helping to explain, I definitely get the process for that particular question now.

@JeffM : I think I understand what you mean in regards to the intuitive notion of a limit at infinity - is c any number that could act as a 'grounder' as such in regards to infinity? So as x -> infinity, and the graphs limit is 1, c could be any number - 10, 100, 1000.. Just something to test how close the graph gets to 1?

And for the last part about the remainder method.. I have no idea how that works. I get how this would be easier to see the part where infinity makes the number too small to be significant however. What is the remainder method? I googled it, but it came up with some stuff that definitely wasn't it.

Thankyou again!
Click to expand...

I may have mislead you. It is not that ANY number will do as c. Only some numbers may do the trick. The point is whether you can find a number, c, so that
F(c) is close enough to a (in calculus the idea of close enough is made more exact) and that F(x) perpetually gets closer to a as x gets bigger than c. F(x) may not be close to a or even getting closer to it when x is a lot less than c. The value of c (and the value of a) are dependent on what you are trying to find the limit of. If there is such an a and such a c for the given function, then the limit of the function as x approaches infinity exists and equals a. Is that clearer?

As for the remainder method, I carelessly used my own personal vocabulary. WHAT I SHOULD HAVE SAID IS polynomial long division. There is an article on it at Wikipedia. See if that helps. Sorry. I shall try to be more careful to avoid my own personal shorthand.
 
@JeffM Definitely clearer. I'm pretty sure I can visualize now :D And thankyou also for clearing up the remainder method, I now understand this also. Definitely some useful information when understanding limits! I have never been able to understand them untill now :)
 
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