Proving continuity of a certain map This problem is quite simple however I am having some troubles to follow what exactly I am supposed to prove here.
Here is the part (a) of this problem which I have solved
For each $\lambda \ge e$ there is a unique $x \in [0,1]$ satisfying $e^x = \lambda x$
The problem is:
Consider the equation $e^x = \lambda x$
Show that the map that takes $\lambda$ to $x$ above is a continuous map on $[e, \infty)$
My question is: Am I supposed to show that the function defined by $\lambda(x) = e^x/x$ is continuous on given domain ? Or is it some other interpretation that I am missing here.
I am having unable to interpret the question correctly, any help would be appreciated.
Edit:
I got that problem actually asks for proving that map $g(\lambda) = x_1$ is continuous where $x_1$ is the unique $x$ satisfying the map $e^x =\lambda x$ . However I am still confused as to how to prove this rigorously?
Any help?
 A: We want to prove that the real function $g:[e,+\infty)\to\mathbb{R}$ such that $g(\lambda)\in[0,1]$ and $e^{g(\lambda)}=g(\lambda)\lambda$ for all $\lambda\geq e$ (which you have already shown to be well defined) is continuous. First of all, let $f:(0,1]\to\mathbb{R}$ be the continuous function $f(x)=e^x/x$. Then, you have probably already shown that $f$ is strictly decreasing on $(0,1]$ because $f'(x)=e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)<0$ for all $x\in(0,1)$. Now, since $f$ is continuous, strictly decreasing, $f(1)=e$ and $\lim_{x\to 0^+}f(x)=+\infty$ you know that $\text{Im}(f)=[e,+\infty)$. Now the proposition that you need to use.

Proposition (Continuity of an inverse function.) If $f:I\to\mathbb{R}$ is a strictly monotonic function (not necessarily continuous) where $I$ is an interval then the function $g:f[I]\to I$ such that $g=f^{-1}$ (which exists since $f$ is injective) is continuous.

Notice that, in our case, the function $g$ of the said proposition is in fact $g:[e,+\infty)\to(0,1]$ such that $\forall\lambda\geq e:e^{g(\lambda)}=g(\lambda)\lambda$ ($g(\lambda)$ can't be $0$ since $e^x$ is never zero). Hence, we conclude that, by this proposition, $g$ is indeed continuous.
The proof of the said proposition is given by @mauro-allegranza on the linked question. However, I will also adapt it for this particular case. Let $\lambda\geq e$ and $\epsilon>0$. If $x:=g(\lambda)\in(0,1)$ (i.e. $\lambda\in(e,+\infty)$) then we know, by the fact that $f$ is strictly decreasing, that $f(x_1)>f(x)=\lambda>f(x_2)$ where $x_1\in(x-\epsilon,x)\cap(0,1]$ and $x_2\in(x,x+\epsilon)\cap(0,1]$ since $x$ is an interior point of $(0,1)$. Now choose $\delta:=\min\{f(x_1)-\lambda,\lambda-f(x_2)\}$. Then, if $\mu\in[e,+\infty)$ is such that $|\lambda-\mu|<\delta$,
$$f(x_1)>\mu>f(x_2)\Rightarrow x_1< g(\mu) < x_2 \Rightarrow |g(\mu)-x|<\epsilon$$
which means that $g$ is continuous on $\lambda\in(e,+\infty)$. For the continuity of $g$ on $\lambda=e$ you can do basically the same by only making use of $x_1\in(x-\epsilon,x)\cap(0,1]$.
