41
$\begingroup$

I learned in grade school that the closer $a$ and $b$ are to one another, the closer $\frac{a}{b}$ is going to be to $1$. For example, $\frac{3}{\pi}$ is pretty close to 1, and $\frac{10^{100}}{42}$ isn't even close to 1.

So, why is:

$$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$$

But:

$$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$$ ?

Seems pretty counterintuitive. What's going on here?

$\endgroup$
2
  • 13
    $\begingroup$ Just because "close together implies ratio is close to $1$" doesn't mean the converse is true. There is zero reason to expect that $a_n/b_n\to 1$ implies even that $a_n-b_n$ is bounded. And the obvious counter-examples prove this. $\endgroup$ Sep 29 '15 at 1:29
  • 7
    $\begingroup$ The relevant meaning of "close" depends on the size of things. $\endgroup$
    – user14972
    Sep 29 '15 at 6:44
105
$\begingroup$

Vadim's answer handles the math (and I've upvoted it), so I will try to provide intuition.

The idea is the word "closer" is relative. That is, in some sense, $100{,}000$ is closer to $100{,}010$ than $1$ is to $0$. Of course, in an absolute sense, $1$ is $1$ away from $0$ while $100{,}000$ is $10$ away from $100{,}010$.

But, say you had $\$1$ and I took away $\$1$ and contrast that with the scenario where you have $\$100{,}010$ and I take $\$10$. In the second scenario, I take more money. But in the first scenario, you care a lot more about the theft.

In your particular example, something similar happens. Yes, $x^2 + x$ and $x^2$ have a very large difference when $x$ is large. But the relative difference is tiny. For example, when $x = 1{,}000$, then $x^2 + x = 1{,}001{,}000$ while $x^2 = 1{,}000{,}000$. So, yes, the difference is large (it's $1000$), but the relative difference is not that big - these two numbers are separated by a mere $0.1\%$ And this percentage shrinks as $x$ gets even bigger.

$\endgroup$
1
  • 15
    $\begingroup$ The analogy explaining that how much you would care if someone took a certain amount of money depends on how much money you have is really helpful to me. Thank you for going out of your way to help me understand this. $\endgroup$ Sep 29 '15 at 1:54
35
$\begingroup$

The reason is that $$\frac{x^2}{x^2+x}=\frac{x^2}{x^2+x}\frac{1/x^2}{1/x^2}=\frac{1}{1+1/x}$$ The fraction is always less than $1$ (for positive $x$), as the denominator is bigger than $1$. However, as $x\to\infty$, we have $\frac{1}{x}\to 0$, so the fraction gets closer and closer to $1$.

$\endgroup$
5
$\begingroup$

For the limit, $$\lim_{x \to \infty} \frac{x^2}{x^2+x}$$

The $x$ term becomes insignificant and $x^2$ is the dominant term as $x$ gets bigger.

$\endgroup$
5
$\begingroup$

$\lim_{x\to\infty} \frac{x^{2}}{x^{2}+x} = 1$

As stated, this is because $x$ grows much more slowly than $x^2$. So the ratio goes to 1.

$1^2\over 1^2+1$$={1\over 1+1}$$={1\over 2}$$=0.5$
$2^2\over 2^2+2$$={2\over 4+2}$$={4\over 6}$$=0.66$
$3^2\over 3^2+3$$={9\over 9+3}$$={9\over 12}$$=0.75$
$10^2\over 10^2+10$$={100\over 100+10}$$={100\over 110}$$=0.91$
$100^2\over 100^2+100$$={10000\over 10000+100}$$={10000\over 10100}$$=0.99$

$\lim_{x\to\infty}[(x^2+x)-(x^2)] = \infty$

This isn't a ratio, so the non-ratio can't go to 1. In fact, it's exactly equal to $x$ for any values:

$\lim_{x\to\infty}[(x^2+x)-(x^2)]$
$=\lim_{x\to\infty}[x^2+x-x^2]$
$=\lim_{x\to\infty}[x^2-x^2+x]$
$=\lim_{x\to\infty}[(x^2-x^2)+x]$
$=\lim_{x\to\infty}[(0)+x]$
$=\lim_{x\to\infty}[x]$
$=\infty$

Obviously, as $x$ goes to infinity, the limit diverges to infinity.

$1^2+1-1^2$$=1+1-1$$=1$
$2^2+2-2^2$$=4+2-4$$=2$
$3^2+3-3^2$$=9+3-9$$=3$
$10^2+10-10^2$$=100+10-100$$=10$
$100^2+100-100^2$$=10000+100-10000$$=100$

$\endgroup$
2
$\begingroup$

Trying to develop an intuitive idea using images and, doing so, cutting a lot of corners.

Suppose Bill Gates gets a raise of 1 dollar an hour. And suppose he decides to work forever. He will make 1$ more every hour. So an infinite amount of money more than without the raise. But do you think this raise is significant? No. The amount of money he makes with or without the raise is almost the same. The ratio between the fortunes he amasses is pretty close to one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.