Let $(f_n):\Bbb R \to \Bbb R$ be a sequence of functions defined by $f_n(x):=\frac{n^2x+n}{n^2+n+1}$ for all $x \in \Bbb R$ and $n \in \Bbb N$. Let $g:\Bbb R \to \Bbb R$ be a function defined by $g(x):=x$. Prove that $(f_n)$ converges pointwise to $g$ on $\Bbb R$.
How to prove it by the $\varepsilon-x$ definition, that is, the definition of converges pointwise? Should I start the proof by divide into some cases, like for $x=0,x=1,x \in (0,1)$, etc?
Edit: Firstly, I compute $|f_n(x)-x|$ as follow: \begin{equation*} \left|\frac{n^2x+n}{n^2+n+1} - x\right| = \left|\frac{n-nx-x}{n^2+n+1}\right|. \end{equation*}
Any idea? Thanks in advanced.
You can rewrite: $$ f_n(x) = \frac{n^2x + n}{n^2 + n + 1} = \frac{x + \frac{1}{n}}{1 + \frac{1}{n} + \frac{1}{n^2}} $$ So it's clear (at least intuitively) that if $n \to \infty$ then $f_n(x) \to x$. To do this rigorously, we have: $$ |f_n(x) - x| = \left|\frac{\left(-\frac{1}{n} - \frac{1}{n^2}x + \frac{1}{n}\right)}{1 + \frac{1}{n} + \frac{1}{n^2}}\right| \leq \left(\frac{1}{n} + \frac{1}{n^2}\right)|x| + \frac{1}{n} \leq \frac{2}{n}|x| + \frac{1}{n} $$ So for any $\epsilon > 0$ let $n$ be large enough so that $\frac{1}{n} < \min\left\{\frac{\epsilon}{4|x|},\frac{1}{n} < \frac{\epsilon}{2}\right\}$. If $x = 0$, then $|f_n(0) - 0| \leq \frac{1}{n}$, so simply let $\frac{1}{n} < \epsilon$.