What is the value of $\displaystyle\lim_{n\to\infty}{\left(n-\left|n\right|\right)}$ 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.
 A: 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.
A: $|n|=n$ 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 $1$.
A: 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. $x=0$.
A: 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.
