Proving this pointwise limit via definition

Question

For $f_n(x) = \frac{nx + x^2}{n^2}$, this converges pointwise to the function $f(x) = 0$.
How would I prove this formally?
This is my attempt:
$$|f_n(x) - f(x)| = |\frac{nx + x^2}{n^2}| \leq |\frac{nx + nx^2}{n^2}| = |\frac{x + x^2}{n}|$$
So if I were to pick $N := \frac{|x + x^2|}{\epsilon}$ then by reversing the steps, we get the required inequality.
My question is: Is my choice of $N$ correct (is there allowance for the absolute values)? And if so, just to confirm for pointwise convergence, the allowance of having $x$ terms in the value for $N$ is O.K. right?

Answer 1

First, $N$ has to be an integer number.

Second, what you wrote $$|f_n(x) - f(x)| = |\frac{nx + x^2}{n^2}| \leq |\frac{nx + nx^2}{n^2}| = |\frac{x + x^2}{n}|$$

is incorrect the middle inequality does not hold necessarly ! For example: if $x=-1$ The RHS is zero, LHS is always positive for all $n \ge2.$

Answer 2

Why did you stop? :) From where you stopped, note that $N:= \lfloor \frac{|x+x^{2}|}{\varepsilon} \rfloor + 1 > \frac{|x+x^{2}|}{\varepsilon}$ and is an integer for sure.

Answer 3

We may establish

$$\left|\frac{nx+x^2}{n^2}\right|<\left|\frac{x}{n}\right|+\left|\frac{x^2}{n^2}\right|<\epsilon.$$

An easy option is to make both terms smaller than $\epsilon/2$, which is achieved by

$$n>\left|\frac{2x}{\epsilon}\right|\text{ and }n>\sqrt{\left|\frac{2x^2}{\epsilon}\right|}$$ and we can take the maximum of the two.