Prove that $f_n(x) = \frac{x^2 + nx + 3}{n}$ converges uniformly over [1,2]

Question

Let $f_n(x) = \frac{x^2 + nx + 3}{n}$, with $f_n$ defined on $[1,2]$. I must show that $\lim_{n \to \infty} f_n$ converges uniformly to $x$ over $[1,2]$.

I start by defining the set

$$ S = \{x : \exists \lim_{n \to \infty} f_n(x)\} $$

Then I define the function

$$ f(x) = \lim_{n \to \infty} f_n(x) = x $$

This seems to prove that $\{ f_n(x)\}$ converges pointwise. Then I try to determine an $\epsilon > 0$, such that

$$ |f_n(x) - f(x)| < \epsilon, \forall x \in S $$

And here I get stuck. I tried finding a maximum value of $|\frac{x^2 + nx + 3}{n} - x|$ using derivatives, but unfortunately this function doesn't have any around $[1,2]$.

Also, I have no idea how should I find a $N_\epsilon$ after finding the $\epsilon$ itself. Any ideas or help would be really appreciated. Thanks!

Answer 1

First note that: $$ \left| \frac{x^2+nx+3}{n}-x\right|=\left|\frac{x^2+nx+3-nx}{n}\right|=\frac{x^2+3}{n}\le \frac{7}{n} $$ So, given $\varepsilon>0$, take $N$ such that $\frac{7}{N}< \varepsilon$. Then $|f_n(x)-f(x)|<\varepsilon$ for every $n\ge N$ and every $x\in [1,2]$.

Answer 2

first we know that $x \in [1,2] \implies (x^2+3) \in [4,7]$

$$|\frac{x^2+nx+3}{n} - x| = |\frac{x^2+3}{n}| < \frac{7}{n} < \epsilon$$

hence $n> \frac{7}{\epsilon}$ suffices hence it converges uniformly