The set \(\{y>0\colon 0\leq y^{p}\lt x\}\) is bounded from above by \( 1 \) if \(x\lt 1\) and by \(x\) if \(x\gt 1\) so it is bounded. Thus \(L:= \sup \{y\in\R\colon 0\leq y^{p}\lt x\} \) is a number: \(L\in\R\). Also \(L\gt 0 \) since \(0^p=0\lt x\). Let us construct a sequence \(y_n\to L\) with \(y_n\in\sup \{y\in\R\colon 0\leq y^{p}\lt x\} \).

This can be done by using that \(L\) is the least upper bound, by choosing \(y_n\in\sup \{y\in\R\colon 0\leq y^{p}\lt x\} \cap (L-1/n,L]\neq \emptyset\) since \(L-1/n\) is not an upper bound for \(\sup \{y\in\R\colon 0\leq y^{p}\lt x\} \). By the sandwich thm, the sequence \(y_n\to L\) since \(\lim_nL-1/n=\lim_n L=L\). Now back to the problem at hand.

Let us show that \(L^p\leq x\) and \(L^p\geq x\). For the former have \(y_n^p\leq x\) and thus \(y_n^p\to L^p\) by the algebra of limits, so \(L^p\leq x\) by closedness of non-strict inequalities.

By contradiction suppose \(x-L^p=: \epsilon_0 \gt 0\). Let us show that \(L\) cannot be an upper bound. We have for \(0\lt \delta\lt 1\) using the binomial expansion

\begin{equation*} (L+\delta)^p = L^p +\sum_{k=1}^p \delta^{k}L^{p-k} \frac{p!}{k!(n-k)!}\leq L^p + \delta \big(p \, p^p \, \max(L,1)^p\big) \end{equation*}

Setting \(\delta=\epsilon_0/ \big(2 p \, p^p \, \max(L,1)^p\big)\) we have that \(L+\delta\in\{y\in\R\colon 0\leq y^{p}\lt x\} \). Contradiction.

This concludes the proof.

We claim that \(\lim a_n=0\).

The inequality above, on the RHS has a weighted combination of previous terms with weights \(1/4\) and \(1/2\) that add up to something smaller than \(1\).

This makes us guess that \(a_n\) has to be dominated by a geometric sequence. Let us try to find \(0\lt q \lt 1\) and \(M\gt 0\) such that \(a_n\leq M q^n\) for all \(n\in\N\). Since \(a_n\geq 0\) and \(Mq^n\to 0\) this allows us to conclude our claim by the sandwich thm.

Let us keep \(M,q\) as parameters and prove the statement \(a_n\leq M q^n\) by induction on \(M\). For \(n=0\) we need to have

\begin{equation*} a_0\leq M \end{equation*}

for \(n=1\) we need

\begin{equation*} a_1\leq Mq \end{equation*}

Cases \(n\geq 2\) are proven by induction. Suppose that \(a_k\leq Mq^k\) for all \(k\leq n\), we need to show that \(a_{n+1}\leq Mq^{n+1}\). We have

\begin{equation*} a_{n+1}\leq \frac{1}{2}a_n + \frac{1}{4} a_{n-1}\leq \frac{M q^{n}}{4} + \frac{M q^{n-1}}{2} \leq^? M q^{n+1} \end{equation*}

where the second inequality holds by induction hypothesis and the last one remains to be shown. In particular the last inequality is equivalent to

\begin{equation*} 4 q^{2}\geq q +2 \end{equation*}

This admits a solution \(q\in(0,1)\) e.g. \(q=9/10\). Taking \(M\geq a_0+a_110/9 \) to satisfy the cases \(n\in\{0,1\}\) concludes the proof.

Let us check the properties of distance functions. For each \(x,y\), \(d^*(x,y)\) is a non-negative real.

It is clear that \(d^*(x,y)=d^*(y,x)\) and \(d^*(x,x)=0\).

If \(d^*(x,y)=0\) we need to show that \(x=y\). This is equivalent to showing that \(x\to \frac{x}{1+|x|}\) is injective. Notice that \(x\) and \(\Big|\frac{x}{1+|x|}\Big|\) have the same sign. Supposing \(x\geq 0\) we can solve for \(t=\Big|\frac{x}{1+|x|}\) to obtain

\begin{equation*} x=\frac{t}{1-t}. \end{equation*}

If \(x\lt 0\) then solve for \(t=\Big|\frac{x}{1+|x|}\) gives

\begin{equation*} x=\frac{t}{1+t}. \end{equation*}

This can be expressed in short by

\begin{equation*} t=\frac{x}{1+|x|} \iff \Big(x= \frac{t}{1-|t|}\text{ and } |t|\lt 1\Big) \end{equation*}

where we notice that \(\Big|\frac{x}{1+|x|}\Big|\lt 1\) for all \(x\in\R\). We have shown the implication \(\implies\) above. The implication \(\Leftarrow\) follows by direct substitution.

Injectivity follows by this explicit inverse computation.

Finally, the triangle inequality can be shown as follows

\begin{equation*} \begin{aligned}[t] d^*(x,z)&=\Big|\frac{x}{1+|x|}-\frac{z}{1+|z|} \Big|=\Big|\frac{x}{1+|x|}-\frac{y}{1+|y|}+ \frac{y}{1+|y|}-\frac{z}{1+|z|} \Big| \\ & \lt \Big|\frac{x}{1+|x|}-\frac{y}{1+|y|}\Big|+\Big| \frac{y}{1+|y|}-\frac{z}{1+|z|} \Big|=d^*(x,y)+d^*(y,z). \end{aligned} \end{equation*}

Let us show that \(x_n=n\) is Cauchy in \((\R,d^*)\). This follows from the fact that for \(m\gt n\) we have

\begin{equation*} d^*(m,n) = \Big|\frac{m}{1+m}-\frac{n}{1+m}\Big| =\frac{m-n}{(1+n)(1+m)}=\frac{m-n}{1+m}\frac{1}{1+n} \leq \frac{1}{n} \end{equation*}

So if \(m\gt n \gt N\) then \(d^*(m,n)\leq \frac{1}{N}\) showing that the sequence is Cauchy.

Now let us show that \(x_n\) does not converge to any \(L\in\R\) i.e. that for any \(L\in\R\) it is not true that

\begin{equation*} \lim_n d^*(x_n,L) =0 \end{equation*}

that is equivalent to showing that

\begin{equation*} \limsup_n d^*(x_n,L) \gt 0. \end{equation*}

Writing out explicitly and using algebra of limits we have

\begin{equation*} \limsup_n \Big|\frac{n}{1+n}--\frac{L}{1+|L|} \Big| = \Big|1--\frac{L}{1+|L|} \Big| \end{equation*}

For the above expression to vanish we would need \(\frac{L}{1+|L|}=1\) but this has no solution on \(\R\) by the above computation of the inverse of \(x\to \frac{x}{1+|x|}\).

We set

\begin{equation*} \begin{aligned}[t] &\bar{d^*}(+\infty,y)=\bar{d^*}(y,+\infty)=\Big|1-\frac{y}{1+|y|}\Big|= \lim_n d^*(n,y) \qquad \forall y\in\R \\ & \bar{d^*}(-\infty,y)=\bar{d^*}(y,-\infty)=\Big|-1-\frac{y}{1+|y|}\Big|= \lim_n d^*(-n,y) \qquad \forall y\in\R \\ & \bar{d^*}(-\infty,+\infty)=\bar{d^*}(+\infty,-\infty)=2= \lim_n d^*(-n,n) \qquad \forall y\in\R \\ & \bar{d^*}(+\infty,+\infty)=\bar{d^*}(-\infty,-\infty)=0 \end{aligned} \end{equation*}

Symmetry and the triangle inequality follow in the same manner as in the first point of this problem.

To show that \(d^*(x,y)=0\;\implies \;x=y\) we discuss: if both \(x,y\in\R\) then it is as in the first point. If both are in \(\{\pm \infty\}\) then it follows from the definition above that eigher both are \(+\infty\) or both are \(-\infty\). If one of them (say \(x\)) is \(\pm \infty\) (say \(+\infty\)) while \(y\in\R\) then the claim follows since

\begin{equation*} \frac{1-\frac{y}{1+|y|}=0 \end{equation*}

has no solution in \(\R\) as shown before. The case \(x=-\infty\) or the case where \(x,y\) are exchanged follow by symmetric reasoning.

Let us show that \((\bar{\R},\bar{d^*})\) is complete. Take \(x_n\in\bar{\R}\) Cauchy, we need to show \(x_n\to L\) for some \(L\in\bar{\R}\). If \(x_n=+\infty\) frequently then let \(x_{n_{k}}\) be a subseq of \(x_n\) such that \(x_{n_{k}}=+\infty\). Clearly \(x_{n_{k}}\to+\infty\) since it is constant but \(x_n\) is Cauchy. A Cauchy sequence with a convergent subseq has to converge to the limit of that subseq (fact from HW01) so \(x_n\to +\infty\). The case \(x_n=-\infty\) frequently is dealt with in the same way.

Finally, if \(x_n\neq\pm \infty\) eventually, we can suppose, by dropping the finitely many terms that \(x_n\neq\pm \infty\) for all \(n\in\N\). Explicitating the Cauchy property we have that for any \(\epsilon\gt 0\), for \(n,m\) large enough it holds that \(d^*(x_n,x_m)\lt \epsilon\) i.e.

\begin{equation*} \Big|\frac{x_n}{1+|x_n|}-\frac{x_m}{1+|x_m|}\Big|\lt \epsilon. \end{equation*}

This means that the sequence \(t_n=\frac{x_n}{1+|x_n|}\) is Cauchy in \(\R\) w.r.t. the standard distance on \(\R\). We have that \(t_n\in[-1,1]\) and since \(\R\) is complete w.r.t. the standard distance we have that \(t_n\to T\) with \(T\in[-1,1]\) i.e.

\begin{equation*} \Big|\frac{x_n}{1+|x_n|}-T\Big|\to 0 \end{equation*}

If \(T\neq \pm 1\) we have that \(|T|\lt 1\) and solving for \(X\) the identity \(\frac{X}{1+|X|}=T\) we obtain \(X\in\R\) so that the above limit

\begin{equation*} \bar{d^*}(x_n,X)\to 0 \end{equation*}

i.e. \(x_n\to^{\bar{d^*}} X\). If \(T=+1\) then the above limit can be rewritten as

\begin{equation*} \bar{d^*}(x_n,+\infty)\to 0 \end{equation*}

i.e. \(x_n\to^{\bar{d^*}} +\infty\). If \(T=-1\) we similarly obtain \(x_n\to^{\bar{d^*}} -\infty\).

Since

\begin{equation*} \big(1+\frac{1}{n}\big)^{n}\lt \big(1+\frac{1}{n}\big)^{n+1} \end{equation*}

and the LHS is decreasing by HW01 we have that

\begin{equation*} \big(1+\frac{1}{n}\big)^{n} \lt 2^3 \end{equation*}

and being increasing the first sequence converges (HW01). We denote

\begin{equation*} \lim_n\big(1+\frac{1}{n}\big)^{n} =e \end{equation*}

By algebra of limits

\begin{equation*} \lim_n\big(1+\frac{1}{n}\big)^{n+1}=\lim_n\big(1+\frac{1}{n}\big)^{n}\lim_n\big(1+\frac{1}{n}\big) =e \; 1 = e \end{equation*}

where the procedure is justified since the limits on the right exist.

Given that \(\big(1+\frac{x}{n}\big)^{n}\) is monotone (HW01), all we need to do is to check that it is bounded from above and below to conclude convergence.

The sequence is positive so \(0\) is a bound from below. If \(x\leq 0 \) then

\begin{equation*} \big(1+\frac{x}{n}\big)^{n}\leq 1. \end{equation*}

It remains to show the case \(x\gt 0\). WLOG \(x\in\N\) since is \(x\gt 0\) and \(x\lt L\) with \(L\in\N\) we have

\begin{equation*} \big(1+\frac{x}{n}\big)^{n}\leq \big(1+\frac{L}{n}\big)^{n} \end{equation*}

so it if sufficient to prove boundedness in \(n\) when \(x\) is a positive integer. Let \(M=M(n)\in\N\) be such that \(M L\leq n \lt (M+1)L\). We have

\begin{equation*} \begin{aligned}[t] & \big(1+\frac{L}{ML}\big)^{(M+1)L}= \big(1+\frac{1}{M}\big)^{(M+1)L} \\ &\leq \Big(\sup_{M}\big(1+\frac{1}{M}\big)^{(M+1)}\Big)^L \leq 8^{L} \end{aligned} \end{equation*}

as required where we used that \(M\to \big(1+\frac{1}{M}\big)^{(M+1)}\) is non-increasing and thus bounded by the value at \(M=1\).

We actually obtained the bound \(e^x\leq 8^{|x|+1}\)

As per the hint and the previous point we have that the limits

\begin{equation*} \begin{aligned}[t] & \lim_{n}\big(1+\frac{x}{n}\big)^{n} \\ & \lim_{n}\big(1+\frac{y}{n}\big)^{n} \\ & \lim_{n}\big(1+\frac{x+y}{n}\big)^{n} \end{aligned} \end{equation*}

exist. So all we need to show is that

\begin{equation*} \lim_{n} \frac{\big(1+\frac{x}{n}\big)^{n} \big(1+\frac{y}{n}\big)^{n}}{\big(1+\frac{x+y}{n}\big)^{n}}=1 \end{equation*}

i.e.

\begin{equation*} \begin{aligned}[t] \lim_{n} \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n}=1. \end{aligned} \end{equation*}

Let us first assume that \(n\gt \max(|x|,|y|,|x+y|)+1000\). This is a limiting procedure in \(n\to +\infty \) so this is valid. We will show that

\begin{equation*} \liminf_{n} \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n}\geq 1 \end{equation*}

and

\begin{equation*} \limsup_{n} \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n}\leq 1. \end{equation*}

Let us show that former:

\begin{equation*} \begin{aligned}[t] \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n} &= \Big(\big(1+\frac{x+y}{n} + \frac{xy}{n^2}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n} \\ & =\Big(1+ \frac{1}{n^2} \frac{xy}{1+\frac{x+y}{n}}\Big)^{n} \end{aligned} \end{equation*}

For \(n\) large enough we have \(\Big| \frac{1}{n^2} \frac{xy}{1+\frac{x+y}{n}}\Big|\lt 1\) since \(\frac{xy}{1+\frac{x+y}{n}}\) is eventually bounded by \(2|xy|\). We can thus apply Bernoulli's ineq to get

\begin{equation*} \begin{aligned}[t] \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n} \geq 1 + \frac{1}{n} \frac{xy}{1+\frac{x+y}{n}} \end{aligned} \end{equation*}

Passing to the \(\liminf\) on both sides we have (using the fact that given two sequences \(a_n\) and \(b_n\) with \(a_n\geq b_n\) we have \(\liminf a_n\geq \liminf b_n \)) that

\begin{equation*} \begin{aligned}[t] \liminf \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n} & \geq \liminf \Big(1 + \frac{1}{n} \frac{xy}{1+\frac{x+y}{n}}\Big) \\ & = \lim \Big(1 + \frac{1}{n} \frac{xy}{1+\frac{x+y}{n}}\Big) = 1 \end{aligned} \end{equation*}

thus showing the first claim.

To show

\begin{equation*} \limsup_{n} \Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n}\leq 1. \end{equation*}

we show that

\begin{equation*} \liminf_{n} \frac{1}{\Big(\big(1+\frac{x}{n}\big)\big(1+\frac{y}{n}\big)\big(1+\frac{x+y}{n}\big)^{-1}\Big)^{n}}\geq 1. \end{equation*}

i.e.

\begin{equation*} \liminf_{n} \Big(\big(1+\frac{x+y}{n}\big) \big(1+\frac{x}{n}\big)^{-1}\big(1+\frac{y}{n}\big)^{-1}\Big)^{n}\geq 1. \end{equation*}

Similar manipulations as before give

\begin{equation*} \begin{aligned}[t] &\Big(\big(1+\frac{x+y}{n}\big) \big(1+\frac{x}{n}\big)^{-1}\big(1+\frac{y}{n}\big)^{-1}\Big)^{n} \\ & = \Big(\big(1+\frac{x+y}{n}\big) \big(1+\frac{x+y}{n} +\frac{xy}{n^2}\big)^{-1}\Big)^{n} \\ & = \Big(1-\frac{1}{n^2} \big(1+\frac{x+y}{n} +\frac{xy}{n^2}\big)^{-1}\Big)^{n} \\ & \geq 1- \frac{1}{n}\big(1+\frac{x+y}{n} +\frac{xy}{n^2}\big)^{-1} \end{aligned} \end{equation*}

where Bernoulli was applied again. Passing to the \(\liminf\) as before gives the result.

Expanding the definition we need to show that

\begin{equation*} \lim_{n\to \infty } \lim_{k\to \infty}\Big(1+\frac{1}{nk} \Big)^k =1 \end{equation*}

in particular showing that the outside limit exists.

We have by Bernoulli that

\begin{equation*} \Big(1+\frac{1}{nk} \Big)^k\geq 1+\frac{1}{n} \end{equation*}

and

\begin{equation*} \Big(1+\frac{1}{nk} \Big)^k = \frac{1}{\Big(1-\frac{1}{nk+1} \Big)^k} \leq\frac{1}{1-\frac{k}{nk+1}} \end{equation*}

so passing to the limit in \(k\) we have

\begin{equation*} 1+\frac{1}{n}\leq \exp(1/n)\leq \frac{1}{1-\frac{1}{n}} \end{equation*}

The conclusion follows by the sandwich theorem.

Above, we have shown that

\begin{equation*} 1+\frac{1}{n}\leq \exp(1/n)\leq \frac{1}{1-\frac{1}{n}} \end{equation*}

so

\begin{equation*} \frac{1}{n}\leq \exp(1/n)-1\leq \frac{1}{n}\frac{1}{1-\frac{1}{n}} \end{equation*}

so

\begin{equation*} 1\leq n\Big(\exp(1/n)-1\Big)\leq \frac{1}{1-\frac{1}{n}}. \end{equation*}

Taking the limit gives the result.