0
$\begingroup$

$$\mathop {\lim }\limits_{x \to {0^ - }} \frac{a}{x}\left\lfloor {\frac{x}{b}} \right\rfloor $$

For sufficently small $x$, $\left\lfloor {\frac{x}{b}} \right\rfloor = - 1$, while $a\over x$ is diverging to $-\infty$.

So, the limit is: $-\infty \cdot (-1)$, Which "equals" to $\infty$.
The problem is I just multiplied a const with infinity and I know it's not defined.

What is the real explanation for the limit diverging to $\infty$?

$\endgroup$

2 Answers 2

1
$\begingroup$

What you have is almost a proof. All you have to do is state things directly in terms of limits of products of functions, rather than in terms of products of the limits themselves. The exact theorem is that if:

$$\begin{align}\lim_{x\to a} f(x)&=L\quad (L > 0) \\ \lim_{x\to a} g(x)&=\infty\end{align}$$

then

$$\lim_{x\to a} f(x)g(x)=\infty$$

The same thing works when $a$ is replaced is $a^-$, $a^+$, $\infty$ and anything else. Of course you have $L<0$ and $g\to-\infty$, so you'll need to spinkle some minus signs on your function in order to get the exact form required by the above theorem (and to justify putting in the minus signs, you need the fact that $\lim$ is linear).

You can also define multiplication of real numbers by infinity in such a way as to make your reasoning correct. Of course, you then need the above theorem in order to prove that the limit of a product is the product of the limits.

$\endgroup$
1
$\begingroup$

As long as $x$ is sufficiently close to $0$, the second factor equals $-1$ as you say, i.e. as long as $-b<x<0$.

The first factor you can make as small as you want (that is, as far from $0$ as you wish), just by letting $x$ get ever closer to $0$. That means that the total product can get as large as you wish. If you phrase that a bit more carefully, you are exactly within the definition of a limit value being $\infty$.

$\endgroup$
2
  • $\begingroup$ Got it. The key-point is showing we can make the expression to be as large as we want (as $x$ goes to $0$). Thanks. $\endgroup$
    – AnnieOK
    Commented May 2, 2014 at 20:08
  • 1
    $\begingroup$ @AnnieOK Almost. In addition to showing that we can make it as big as we want, we also need to ensure that it doesn't go back down again. Usually, when showing that $\lim_{x \to a}f(x) = \infty$ for some $f$ and $a$, this is done by proving that for any potential lower bound $K$, as long as $x$ is close enough to $a$, $f(x) > K$. In textbooks (and actual proofs) the usual phrasing is along these lines: "$\lim_{x \to a} f(x) = \infty$ iff for any real number $K$ there is a $\delta > 0$ such that for any $x$ with $|x-a| < \delta$ (this is the 'close enough'-condition) then $f(x) > K$". $\endgroup$
    – Arthur
    Commented May 3, 2014 at 11:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .