Determine for what values of x the sequence fn(x) = (nx+5)/(6n+7) converges pointwise

Question

May you tell me if this reasoning is correct? Thank you so much!

fn(x) = (nx+5)/(6n+7)

Determine for what values of x the sequence converges pointwise.

The limit function f(x) = x/6

D(x)= Absolute value of (fn(x) - f(x))

= Absolute value of ((nx+5)/(6n+7) - x/6)

= Absolute value of (30-7x)/(36n+42)

The limit of D(x) when n goes to infinite is 0, for all values of x.

D(x) is the distance function between fn(x) and f(x).

Thus fn(x) converges uniformly for all values of R, and for that reason it converges pointwise for all R.

Answer 1

$(f_n)_{n \in \mathbb{N}}$ is pointwise convergence, your proof is correct. An easier way is the following: $$f_n(x) = \frac{n(x+5/n)}{n(6+7/n)} = \frac{x+5/n}{6+7/n} \to \frac{x}{6} \ \ (n \to \infty)$$ But the sequence is not uniformly convergent: Let $\epsilon > 0$ and $N \in \mathbb{N}$. Then for $x=-36N-42$ and $n \geq N$ we have

$$f_n(x)-f(x) = \frac{30-7(-36N-42)}{36n+42} = \frac{30}{36n+42}+7 > 7$$

Thus for e.g. $\epsilon=1$, the condition "$\forall \epsilon>0 \ \exists N \in \mathbb{N}$ s.t. $|f_n(x)-f(x)| < \epsilon \ \forall x \in \mathbb{R}$" is not true and $f_n$ is not uniformly convergent.