Prove $\{h_n\}_{n=1}^\infty$ converges uniformly on $[a,1]$

Question

Let $h_n: [a,1] \to R$ defined by:

$h_n(x) = n^2x$ for $a \le x < \frac 1n$ and $h_n(x) = \frac 1x$ for $ \frac 1n \le x \le 1$.

Show that $\{h_n\}_{n=1}^\infty$ converges uniformly on $[a,1]$ where $0 < a < 1$.

I guess my main concern here is what function $h(x)$ does $\{h_n\}_{n=1}^\infty$ converge pointwise to? I think its $h(x) = \frac 1x$ but I'm not sure. After I figure that out I'm pretty sure I can satisfy the necessary epsilon inequality for uniform convergence.

Thanks for the help!

Answer 1

Take $n_0>1/a$. For all $n\ge n_0$, $x\in[a,1]$: $1/n<1/n_0<a\le x$, i.e, you are in the second case and $f_n(x)=1/x$. And uniform convergence $\implies$ pointwise convergence to tha same limit.

Answer 2

Your guess is correct: the sequence of functions converges pointwise to $h(x) = 1/x$ on any interval $[a, 1]$ with $0 < a < 1$. This is because $1/n$ eventually gets smaller than $a$, so for this $n$ (and all subsequent values of $n$), $h_n(x) = 1/x$ because $1/n < a \le x \le 1$.