1. ## Question about limits

In part (a) of my questions I'm asked to show $\displaystyle \lim_{n \rightarrow{\infty}}[\frac{1}{n-1}]^\frac{1}{n} = 1$

I have done this, and part (b) asks me to consider a sequence of functions

$\displaystyle f_n : [0,1] \to \mathbb{R} : x \to \frac{x}{1 + x^n}$,

then for each fixed $\displaystyle n \in \mathbb{N}$ determine the point $\displaystyle \varepsilon(n) \in [0,1]$ at which the function $\displaystyle f_n$ attains its maximum value and hence, calculate

$\displaystyle \lim_{n \rightarrow{\infty}}f_n(\varepsilon(n))$

I'm told I may use part a. I'm not sure if $\displaystyle \varepsilon(n)$ is meant to represent x values in which case the max would be achieved (as I can see) at x = 1, but I can't see the relation with part (a) by doing that.

2. Originally Posted by pineapple89
In part (a) of my questions I'm asked to show $\displaystyle \lim_{n \rightarrow{\infty}}[\frac{1}{n-1}]^\frac{1}{n} = 1$

I have done this, and part (b) asks me to consider a sequence of functions

$\displaystyle f_n : [0,1] \to \mathbb{R} : x \to \frac{x}{1 + x^n}$,

then for each fixed $\displaystyle n \in \mathbb{N}$ determine the point $\displaystyle \varepsilon(n) \in [0,1]$ at which the function $\displaystyle f_n$ attains its maximum value and hence, calculate

$\displaystyle \lim_{n \rightarrow{\infty}}f_n(\varepsilon(n))$

I'm told I may use part a. I'm not sure if $\displaystyle \varepsilon(n)$ is meant to represent x values in which case the max would be achieved (as I can see) at x = 1, but I can't see the relation with part (a) by doing that.
Let's solve $\displaystyle f_n'(x)=0 \implies \frac{1-(n-1)x^n}{(x^n+1)^2}=0 \implies 1-(n-1)x^n=0 \implies x=\left(\frac{1}{n-1}\right)^\frac{1}{n}$.

Hence $\displaystyle f_n(x)$ has a critical point at $\displaystyle x=\left(\frac{1}{n-1}\right)^\frac{1}{n}$. Now verify that this is indeed a maximum.

3. Aah, I see, I was thinking in terms of looking for the supremum like when establishing uniform convergence. Didn't think of maximum like that. Silly me Thanks!