# HW 8: Challenge 3(a,b)

Consider the sequence of functions $g_n:[0,\infty]\to R$ defined by $g_n(x):=\frac{x^n}{n}$

a) Prove that $(g_n)$ converges uniformly on $[0,1]$ and find $g=\lim g_n$.

$(g_n)=(\frac{x^n}{n})=\{x, \frac{x^2}{2},\frac{x^3}{3},\hdots\frac{x^n}{n}\}$

In order to better examine this, look at $x=.5$, then $(g_n)=(\frac{.5^n}{n})=\{.5, \frac{.5^2}{2},\frac{.5^3}{3},\hdots\frac{.5^n}{n}\}$. The limit of $\frac{.5^n}{n}$ as $n\to\infty$ is 0. This holds similarly for $x\in(0,1)$

For $x=0, 0^n/n=0\;\forall n$ and $x=1, 1^n/n=1/n\to 0$, so $(g_n)$ converges pointwise to $g(x)=0=\lim g_n$

Let $\epsilon>0$ be arbitrary and choose $\frac{1}{N}<\epsilon$ (which exists by Archimedean Property). Let $n\geq N, x\in[0,1]$ both be arbitrary.

Examine $|g_n(x)-g(x)|=|g_n(x)-0|=|\frac{x^n}{n}|$. Notice that $|\frac{x^n}{n}|\leq \frac{1}{n}\leq \frac{1}{N} <\epsilon$, since $n\geq N$. Thus, since $\epsilon, n, \text{ and } x$ were arbitrary (within the stated constraints), for all $\epsilon>0,\exists N>\frac{1}{\epsilon}\in N$ such that $g_n(x)-g(x)|<\epsilon$ whenever $n \geq N$ and $x\in [0,1]$. Thus by definition of unform converges, $(g_n)$ converges uniformly on $[0,1]$.

b) Show that $g=\lim g_n$ is differentiable on $[0,1]$ and compute $g'(x)$.

As shown in part a) $g=\lim g_n=0$.

Let $c\in[0,1]$ be arbitrary and consider $\lim_{x\to c} \frac{g(x)-g(c)}{x-c}$. Since $g$ is constant and defined for $c$, this is the same as $\lim_{x\to c} \frac{0-0}{x-c}=0$. This however, if the definition of the derivative of g at c, which therefore exists. Since $c$ was arbitrary, the derivative $g'(x)=0$ exists for all points in $[0,1]$, and thus by definition of differentiable,$g'(x)=0$ is differentiable on $[0,1]$.

#### 2 Responses

1. Gregory Bischoff says:

All around very good! Structure is nice and clear. I’d only say to be a bit careful with typesetting. Also, when showing the existence of the limit in part b, I think you’d need to account for the fact that the denominator is 0 when x = c. You could do this by invoking L’Hopital’s rule. (though I feel I may not be right there as it is ALWAYS zero otherwise…)