Limits of $f(x)=x-x$

Question

It's obvious that $f(x)=x-x=0$. But what exactly happens here?

You have a function $f(x)=x-x$ and you have to calculate the limits when $x\to \infty$

This'll be like this: $$\lim\limits_{x\to \infty}f(x)=\infty - \infty$$ That's an indetermination, and you have to multiply both sides with the conjugate of $f(x)$, which is equal to $x+x$.

\begin{align} f(x)&=x-x\\ &=\frac{(x-x)(x+x)}{x+x}\\ &=\frac{x^2-x^2}{x+x} \end{align}

If we do the limits now the answer is going to be: $$\lim\limits_{x\to \infty}f(x)=\frac{\infty - \infty}{\infty+\infty}$$ Which it's another type of indetermination(I think).

What happens here?Can there be an error multiplying with it's conjugate in both sides? Is there another case like this? Or am I completely wrong?

Answer 1

$\lim\limits_{x\to\infty}(x-x)$ isn't indeterminate: since $x-x=0$ for all $x$, it's simply $\lim\limits_{x\to\infty}0=0$. You certainly don't have to multiply by $\frac{x+x}{x+x}$, since there's a very simple, direct way to evaluate the limit. If you do perform this unnecessarly multiplication to get $$\lim_{x\to\infty}\frac{x^2-x^2}{x+x}\;,$$ you get a second chance to realize that you can simplify the expression: $\frac{x^2-x^2}{x+x}$ is identically $0$ for $x\ne 0$, and the singularity at $x=0$ is irrelevant to the limit as $x\to\infty$, so once again you have simply $\lim\limits_{x\to\infty}0=0$.

Answer 2

Never treat the symbol $\infty$ as a number subject to the usual rules of arithmetic. In most limit questions that you are asked, that procedure will give you the wrong answer. It can be useful, in a question about $\lim_{x\to \infty}f(x)$, to imagine that $x$ is a very large specific number, like $10^{40}$. It is clear that if $x=10^{40}$, then $x-x=0$.

Answer 3

The problem with indeterminate forms is that you measure things which approach the same limit, but at possibly different rates and what you are interested is the limit of the difference.

This is why $\lim 2n-n=\infty-\infty$ is infinite, because we evaluate the limit of the difference, not the difference of limits.

On the other hand, $f(x)=x-x$ is constantly zero, the difference is constant and therefore the limit is constant too, similarly $g(x)=\frac{x}{2x}$ has a constant ratio of $\frac12$ when approaching infinity, and the same logic applies.

Answer 4

The key point which is at stake here is that limits are NOT linear when they do not exist (or are infinity). In other words, $\lim_{x\rightarrow\infty} (x-x)$ does NOT equal $\lim_{x\rightarrow\infty}x-\lim_{x\rightarrow\infty}x$, whereas $\lim_{x\rightarrow\infty}(1/x-1/x)=\lim_{x\rightarrow\infty}(1/x)-\lim_{x\rightarrow\infty}(1/x)=0-0=0$ is valid since the individual limits exist and are finite. In particular, it is extremely dangerous to even hypothetically reason that $\lim_{x\rightarrow\infty}=\infty-\infty$ not only because the right hand side makes no sense but also because this masks the subtle lower order behavior that goes on. Take for example $\lim_{x\rightarrow\infty} (1+x-x)$ and erroneously conclude that it's 0 by thinking $\lim_{x\rightarrow\infty}(1+x)=\infty$ and $\lim_{x\rightarrow}(-x)=-\infty$ whereas the correct answer is simply 1. I sincerely urge you to drop this handwavy plugging in of $\infty$ into limits and use more grounded techniques to reason them out (at least until you are comfortable with the rules at work here).

Answer 5

The best way of doing limits is by simplifying from the start, so you only have to do this: \begin{align}f(x)&=x-x\\&=0\end{align}With this, you know that whatever the value of $x$ is, $f(x)$ is going to be $0$.

Answer 6

The problem here is that symbol manipulation is being substituted for reasoning. The limit does not mean that we can simply instantiate the $x$ variable as the $\infty$ symbol and do some arithmetic. The reason that $\infty - \infty$ is indeterminate is that $\infty$ is not a specific number, but a concept. You've substituted a not-a-number for a variable and then concluded that the formula is indeterminate.

One way to rescue this is this: since you have substituted some concrete entity for $x$, both instances of that entity have to be the same manifestation of that entity. That is to say, your $\infty - \infty$ is not just any two infinities; it is the same infinity occurring twice: that infinity which was substituted for $x$. This infinity has to be equal to itself.

But of course there can be two different infinities in a formula obtained differently. For instance suppose we have $x - y$, and both $x$ and $y$ tend to infinity. Since they are independent, they go to different infinities, and so if we substitute, we get an indeterminate $\infty - \infty$. Here, each infinity is from a different substitution and so a different infinity object.

Our mathematical typography does not visually distinguish two infinities that are multiple occurences of the same infinity, and two infinities of independent origin. Both just look like $\infty$.

Ultimately, the limit concept is about approaching but not reaching. To ask whether there is a limit as $x\to\infty$ really means "does the formula converge on a value as $x$ gets arbitrarily large". There is no question of substituting some concrete infinity for $x$; it is a question of probing the space of $x$ to large values to see whether the difference between one formula and another (its supposed limit) can be shown to keep diminishing as $x$ moves toward the limit.

It is because limit means "approach, but do not reach", we can evaluate this:

$$\lim_{x\to 0} {x\over x}$$

The function $x\over x$ is exactly like $1$ except that it is not continuous over $x = 0$. It has no value there. Yet, the limit of this function as $x$ approaches zero (from either side) is 1.

We cannot calculate this limit by substituting $0$ for $x$, similarly to the difficulty of substituting $\infty$ for $x$ in a $lim_{x\to \infty}$ situation.

We can sometimes short-cut to a limit by substituting zero: in cases where we do not get an absurdity like division by zero.

Answer 7

I think of limits in terms of epsilons and deltas. For this problem, if $\epsilon$>0, then there does exist an r in R (choose any real number you desire) such that for all x>r in the domain, $\left|f(x)-0\right|=\left|(x-x)-0\right|=\left|0-0\right|=0<\epsilon$.

I've never evaluated a limit by plugging a value in in my life; so I have no problem with this problem.