Let $h_n : [0,1] \rightarrow \mathbb{R}$ Such that $h_n(x) = 1 − nx$ if $0 \leq x \leq \frac{1}{n}$ and $h_n(x) = 0$ if $\frac{1}{n}$ if $\frac{1}{n} \leq x \leq 1.$ Determine if $h_n$ converges uniformely
In this case, we note that it doesn't converge uniformly since the $lim_{n\rightarrow\infty} 1-nx = 1$ when $0 \leq x \leq \frac{1}{n}$, but $lim_{n\rightarrow\infty} 0 = 0$ when $\frac{1}{n} \leq x \leq 1$ Hence we have two different limits. Is my reasoning correct?
As explained in comments the statement $\lim_{n \to \infty} (1 - nx ) =1 $ for $0 \leqslant x \leqslant 1/n$ is specious.
A valid argument is, for fixed $x \in (0,1]$ and $N$ such that $1/N < x$, we have $h_n(x)=0$ for all $n \geqslant N$. For $x=0$ we have $h_n(0) = 1$. Thus $h_n$ converges pointwise to $h$ where $h(0) =1$ and $h(x) = 0$ for $0 < x \leqslant 1$. Since $h$ is discontinuous and each $h_n$ is continuous, convergence cannot be uniform.
Alternatively, after establishing that the pointwise limit is $h(x)$ (above), we have
$$|h_n(x) - h(x)| = \begin{cases}0, & x = 0 \\ |1 - nx|, & 0 < x \leqslant 1 \end{cases}$$
Thus, for $n> 1$ we have $\sup_{x \in [0,1]}|h_n(x) - h(x)| = n -1$. Since the limit of this supremum is not $0$, convergence is not uniform.