Question: I saw this definition in one of the computer science book but I am unable to recall the theorem name. Can someone please provide the reference?

$$\lim\_{n \to \infty} \frac{f(n)}{g(n)} = c > 0$$ means there is some $n\_{0}$ beyond which the ratio is always between $\frac{1}{2}c$ and $2c$.

Answer: 
Since $\|A\| = 1$, using the Neumann series we can indeed invert $A - \lambda I$ when $|\lambda|>1$. For $|\lambda|<1$, the sequence $x\_n = \lambda^n$ lies in the kernel of $A-\lambda I$ and it is not invertible.

Then the conclusion follows by recalling that the spectrum of a bounded operator is always closed.
First note that $A$ is an isometry, so $\|A\| = 1$. Hence $\sigma(A)$ lies in the closed unit ball of the complex plane $C$.

Define the adjoint of $A : X \rightarrow Y$ to be $A^\* : Y^\* \rightarrow X^\*$ such that $A^\*(\alpha)(x) = \alpha(Ax)$ where $\alpha \in Y^\*$ and $x \in X$ (here $X,Y$ are Banach spaces). Then $\sigma(A) = \sigma(A^\*)$. Then easy computation should give you the adjoint of left shift is the right shift (D'oh!). It is very easy to see that the spectrum of the right shift is all the closed unit disc by considering $x = (1,\lambda, \lambda^2, \lambda^3,...)$.