I think I am close to solving the problem, but I hit a dead end trying to show that $\lambda_n$ converges to $(n+\frac{1}{2})^2\pi^2$ as $n\to\infty$. Specifically, how can I re-express $\tan\mu=\mu$ in terms of $n$ and $\pi$ s.t. $\lambda=\mu^2$?
Show that $\lambda=\mu^2$ is an eigenvalue of the BVP
$$\begin{cases} -u''(x)=\lambda u(x),\quad 0<x<1 \\ u(0)=0,\quad u(1)=u'(1), \end{cases} $$
provided that $\mu$ is a root of the equation $\tan\mu=\mu$. By considering the graphs of $\tan\mu$ and $\mu$, show that there are an infinite number of eigenvalues $\lambda_n$ and that $\lambda_n\to(n+\frac{1}{2})^2\pi^2$ as $n\to\infty$.
Solution:
Consider three cases, $\lambda>0, \lambda=0,\ \text{and}\ \lambda<0$.
Case I:
Suppose $\lambda<0$. We write $\lambda=-\mu^2$, which gives us the characteristic equation $r^2+\lambda=0\to r^2-\mu^2=0$. So we have the general solution $$u(x)=Ae^{\mu x}+Be^{-\mu x}.$$
The first BC implies that $$Ae^{0}+Be^{0}=0\implies A+B=0\implies B=-A\implies u(x)=A(e^{\mu}-e^{-\mu}).$$
The second BC implies that $$A(e^{\mu}-e^{-\mu})=A\mu(e^{\mu}-e^{-\mu}),$$ which can be expressed as $2A\sinh\mu=2A\mu\cosh\mu$. So $\mu=\tanh\mu$. But there exists no $\mu\neq0$ such that $\mu=\tanh\mu$.
Case II:
Suppose $\lambda=0$. Then $u''(x)=0$ which has the general solution $u(x)=A+Bx$.
The first BC implies that $A=0$, and the second BC implies that $B=B$. Therefore $\lambda=0$ is an eigenvalue of the BVP.
Case III:
Suppose $\lambda>0$. We write $\lambda=\mu^2$, which give us the characteristic equation $r^2+\lambda=0\to r^2+\mu^2=0$. So we have the general solution $$u(x)=A\cos\mu x+B\sin\mu x.$$
The first BC implies that $$A\cos0=0\implies A=0\implies u(x)=B\sin\mu x.$$
The second BC implies that $$B\sin\mu=B\mu\cos\mu.$$ So indeed $$\mu=\tan\mu.$$
By considering the graphs of $\tan\mu$ and $\mu$, we see that there are an infinite number of eigenvalues $\lambda_n$.
This is most easily seen visually: The roots of $\mu=\tan\mu$ are the points of intersection of the line $y=x$ and $y=\tan x$. So you're essentially finished; just note that the branches of the function $\tan\mu$ lie near their asymptotes at $\mu=\pi(n+1/2)$ at these points.
You cannot solve that equation explicitly (it's transcendental), so the best you can do is provide the asymptotic behaviour: $\mu_n\sim \pi(n+1/2)$ as $n\to\infty$ (this notation also addresses the comments in the question about the limit notation not being accurate).