What's the real result of $\lim_{x\to \infty} x - x$

Question

So In a practice problem, I've found some complicated lim to compute and at the end I got this : $$\lim_{x\to \infty} x - x$$

My answer was that it was equal to $0$ , because in my reasoning $x-x = 0$ , But when I asked a math teacher I knew he told me that it was an indeterminated form because as we knows the limits of $\infty - \infty$ is undefined . While my class teacher told me that it was correct, Both teacher are old and very good and I don't know which one is right, because for me bot explanations seems logical, so please let me know what's the right answer and why ?

Also does this fact apply on those limits :

$$\lim_{x \to \infty} \frac{x}{x} $$ $\frac{\infty}{\infty}$ is an undefined limits

$$ \lim_{x \to \infty} ax - x $$

Where $a$ is a known real number

First I thought that I was good at computing limits but this case really confuse me now !

Answer 1

You are entirely correct that $$\lim_{x\to\infty}(x-x)=0.$$ Since $x-x=0$ for all $x$, as $x$ approaches $\infty$ the values of $x-x$ approach $0$ (they are in fact always exactly $0$).

An "indeterminate form" is just a general form of a limit such that you can't say what the limit is automatically based on its form. So for instance, a limit of the form $$\lim_{x\to a}(f(x)-g(x))$$ where $f(x)$ and $g(x)$ are both going to $\infty$ is "indeterminate", because this alone is not enough information to determine the limit. In your example, the limit was $0$. But in the example $$\lim_{x\to\infty}((x+1)-x),$$ the limit would be $1$ instead, since $(x+1)-x=1$ for all $x$. Or if you had $$\lim_{x\to\infty} (2x-x),$$ the limit would not exist (it diverges to $\infty$), since $2x-x=x$ and $x$ is going to $\infty$.

So in any particular example, a limit of "indeterminate form" does have a definite value (or the limit may not exist). It's just that you can't tell in advance what the limit will be merely from the fact that it looks like "$\infty-\infty$".

Answer 2

$$\lim_{x\to \infty}x-x $$ is an indeterminate form but it is zero in this particular case.

$$\lim_{x\to\infty}\frac xx $$ is an indeterminate form but it is $1$ in this case.

$$\lim_{x\to+\infty}(ax-x)=(a-1)\lim_{x\to+\infty}x$$

$=+\infty$ if $\;a>1$ and $\;-\infty$ if $a<1$.

Answer 3

For me: the real question stated would be: What is the difference between the function $f(x)=x$ and a function $g(x)=x$ when $x$ goes to infinity.

As the functions are equal for equal values of $x$, this simplifies to $$ \lim_{x \to \infty} 0 = 0$$

So, as functions are equal, there is no indeterminate form.

Answer 4

The answer to the first one is $0$. The logic that $\lim_{x\to\infty}(x-x)=\infty-\infty$ presumes the addition property of limits holds in this case, when it doesn't. We can say that $\lim_{x\to\infty}(f+g)(x)=\lim_{x\to\infty}f(x)+\lim_{x\to\infty}g(x)$, provided the two limits exist. In this case, where $f(x)=x$ and $g(x)=-x$, the individual limits as $x$ goes to $\infty$ do not exist, so we cannot use the addition property. So instead we have $\lim_{x\to\infty}(x-x)=\lim_{x\to\infty}(0)=0$.

Similarly, for $\frac{\lim_{x\to\infty}f(x)}{\lim_{x\to\infty}g(x)}=\lim_{x\to\infty}(\frac{f(x)}{g(x)})$, the individual limits must exist, and of course the denominator must be nonzero. So we actually have $\lim_{x\to\infty}(\frac{x}{x})=\lim_{x\to\infty}(1)=1$.

For $\lim_{x\to\infty}(ax-x)=\lim_{x\to\infty}(x(a-1))=(a-1)\lim_{x\to\infty}x$, we see that the limit exists only when $a=1$, and the limit is $0$ in that case.

Answer 5

What is denoted by $$\infty-\infty$$ is indeed a so-called indeterminate form, which can take different values depending on the particular ways to reach the two infinities. Informally, one could express the indeterminate case in your example as

$$\left.(x-x)\right|_{x=\infty}$$ i.e. the difference evaluated at $x=\infty$, though this is not "allowed" in the reals as $\infty\notin\mathbb R$.

But the expression

$$\lim_{x\to\infty}(x-x)$$ is not an indeterminate form, it is the real $0$, rigorously obtained from the rules of limit calculation.

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Similarly,

$$\left.\frac xx\right|_{x=\infty}=\frac{\infty}{\infty}$$ denotes an indeterminate form, and

$$\lim_{x\to\infty}\frac xx=1$$ is the limit of this particular indeterminate form, which is a well-defined real.

By contrast, $\dfrac x{x^2}$ and $\dfrac{x^2}x$ generate two $\dfrac\infty\infty$ forms but

$$\lim_{x\to\infty}\frac x{x^2}=0,$$ while $$\lim_{x\to\infty}\frac{x^2}x$$ does not exist.