I am a beginner to calculus, and I have a simple question on limits.

Consider the function $f(x)= 1/x$ for all real $x$. Then we know that upper limit of $x$ tends to infinity is $0$. This is because x can be arbitrarily large, and the larger $x$ is, the smaller $f(x)$ is. So $f(x)$ gets closer and closer to $0$, but it never is equal to $0$. So upper limit of $x$ tends to infinity is $0$. But we can also think of it this way that as $x$ increases, $f(x)$ decreases. As $f(x)$ tends to $0$, we can also say that it tends towards $-1$ because $x$ is getting larger and larger, thus $f(x)$ is getting smaller and smaller, and the smaller it gets, the closer it gets to both $0$ and $-1$. My question is: Why do we say upper limit of $x$ tends to infinity is $0$ and not $-1$?

  • 1
    $\begingroup$ Because -1 is always at a distance of at least 1 from $f(x)$, whereas $0$ is at distance $\varepsilon$ if $x$ is big enough. $\endgroup$ – Andy Apr 23 '13 at 18:27
  • $\begingroup$ The function must not just get closer and closer to the limit, but must be able to get arbitrarily close. $f(x)$ can get closer and closer to $-1$, but you cannot make $f(x)$ as close as you like to $-1$. However not only does $f(x)$ get closer and closer to $0$, you can make $f(x)$ as close as you like to $0$ by taking $x$ sufficiently large. $\endgroup$ – user50229 Apr 23 '13 at 18:58

I think you are confusing limit with lower bound. A lower bound is a number below all the values of the function. $-1$ is a fine lower bound for $\frac 1x$ as long as $x \gt 0$ because it is less than all the values of $\frac 1x$. $-\frac 12,-2,$ and $-\pi$ are also lower bounds for the same reason. A limit, if it exists, is unique. It is what the function "comes very close to". $\lim_{x \to \infty} \frac 1x = 0$ because, as you say, for $x$ very large $\frac 1x $ is very close to zero. It is not very close to $-1$, so $-1$ is not a (or the) limit of $\frac 1x$


$f(x)\underset{x\to+\infty}{\longrightarrow} l \underset{definition}{\Leftrightarrow} \forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l\right|\le\varepsilon$

For all $\varepsilon > 0$, we can take $A=\cfrac{1}{\varepsilon}$ and then, $\forall x \ge A,\left|\cfrac{1}{x}-0\right|=\cfrac{1}{x} \le \cfrac{1}{A}=\varepsilon$

So $\boxed{\cfrac{1}{x}\underset{x\to+\infty}{\longrightarrow}0}$

Now take $\varepsilon = \cfrac{1}{2}$

$\forall A \ge 0,$ for $x=2+A$, $\left|\cfrac{1}{x}-1\right|=\left|\cfrac{1}{2+A}-1\right|=\left|\cfrac{-1-A}{2+A}\right|=\cfrac{1+A}{2+A}\ge\cfrac{1}{2}=\varepsilon$

So since it doesn't work for that specific $\varepsilon$, it can not work $\forall \varepsilon$ so


(for some reason I can't get the strike on the arrow... If someone knows how to format it properly, please edit)

Also, the limit is always unique.

Suppose $f(x)\underset{x\to+\infty}{\longrightarrow} l_1$ and $f(x)\underset{x\to+\infty}{\longrightarrow} l_2$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l_1\right|\le\varepsilon$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l_2\right|\le\varepsilon$

Since $\forall x,\left|l_1-l_2\right|=\left|(l_1-f(x))-(l_2-f(x))\right|\le \left|f(x)-l_1\right|+\left|f(x)-l_2\right|$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|l_1-l_2\right|\le \varepsilon$

We can remove useless variables from the formula and we get

$\forall \varepsilon >0, \left|l_1-l_2\right|\le \varepsilon$

So $\left|l_1-l_2\right|=0$

That is $\boxed{l_1=l_2}$

  • $\begingroup$ Could you please explain it? $\endgroup$ – abcd Apr 23 '13 at 18:28
  • $\begingroup$ @abcd: I edited. $\endgroup$ – xavierm02 Apr 23 '13 at 18:50

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