What is the value of $\displaystyle\lim_{n\to\infty}{\left(n-\left|n\right|\right)}$

Question

Let $$x=\displaystyle\lim_{n\to\infty}{\left(n-\left|n\right|\right)}$$ It is obviously that $x=0$, but is it really simple? Can we just say that $|n|=n$ at $n\to\infty$ because $n>0$?
For examle, let $\displaystyle{y}=\lim_{n\to0}\frac{n^2+n}{n}$. We can see that $y=1$, but, if we write $n^2=n$ because it is true at $n=0$, then $y=2$, so we cannot write this. For any $\displaystyle\lim_{n\to{a}}f(n)$ we know that $f(n)$ do not depends on $a$. If this is true, how can we prove that $x=0$?
I asked Wolfram Alpha for this problem. His step-by-step solution is very complicated and I could not understand it.

Answer 1

Yes you can say that. You could prove it using the definition of $\displaystyle\lim_{n\to\infty}$:

We say that $\displaystyle\lim_{n\to\infty} (n-|n|)= x$ if $\forall \varepsilon > 0\ \exists K$ so that if $n > K$ then $|(n-|n|)-x| < \varepsilon$. We can prove that $x = 0$ since for any $\varepsilon > 0$ you can choose $K = 0$ and the definition holds.

Answer 2

$|n|=n$ is true for all positive $n$, so that $|n|-n=0$ holds in the neighborhood of $\infty$, and for this reason the limit is $0$.

$n^2=n$ is true for $n=0$ only, so you can't use this relation in the neighborhood of $0$. By contrast, $\frac{n^2+n}{n}=n+1$ holds in such a neighborhood, and the limit is $1$.

Answer 3

The staement $$x=\lim_{n\to\infty}(n-|n|)$$ is equivalent to $$ \forall \epsilon>0\colon\exists n_0\in\mathbb N\colon\forall n>n_0\colon |(n-|n|)-x|<\epsilon.$$ As $n-|n|=0$ for $n>0$ (and $n>0$ holds for almost all - in fact for all $n\in\mathbb N$), this simplifies to $|x|<\epsilon$ for all $\epsilon$, i.e. $x=0$.

Answer 4

Definition of Absolute Value \[ |x|=\left\{ \begin{array}{cc} x & : x\ge 0 \\ -x & : x<0 \\ \end{array} \right. \] Using this definition, we can answer the first two questions \[ \lim_{n \to \infty} \left(n-|n|\right) \] Note that $|n|=n$ as $n$ approaches $\infty$ because as $n$ approaches $\infty$, $n$ is positive. So by using the definition of absolute value, we can write \[ \lim_{n \to \infty} \left(n-|n|\right)= \lim_{n \to \infty} \left(n-n\right)= \lim_{n \to \infty} 0=0 \] Also the notation $n \to \infty$, is read "as $n$ approaches $\infty$". Not "at $n$ approaches $\infty$". That one word difference can totally curb your understanding of limits. \[ \lim_{n \to 0} \frac{n^{2}+n}{n} \] Although it's true that $n^{2}=n$ at $n=0$, this is not true as $n$ approaches $0$.

For example, here's a possible value of $n$ as $n$ approaches $0$ from the left \[ (-0.001)^{2}\ne -0.001 \] And now from the right \[ 0.001^{2}\ne 0.001 \] To evaluate this limit, we must take it out of its current indeterminate form \[ \lim_{n \to 0} \frac{n^{2}+n}{n}= \lim_{n \to 0} (n+1)=0+1=1 \] Now we can see that this limit will never equal 2.

Knowing how to properly read the notation is critical to understanding.