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Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places). lim e5t - 1 / t, t = ± 0.5, ± 0.1, ± 0.01, ±0.001, ±0.0001 t-->0

The fixed points of a Mobius transformation $f(z)=\frac{a z+b}{c z+d} \quad(a d-b c \neq 0)$ are those points z for which $f(z)=z$ . Let $D=(d-a)^{2}+4 b c$ . a) Determine all fixed points when c = 0. b) If $c \neq 0$ and $D \neq 0$ , prove that f has exactly 2 fixed points $z_{1}$ and $z_{2}$ (both finite) and that they satisfy the equation $\frac{f(z)-z_{1}}{f(z)-z_{2}}=R e^{i \theta} \frac{z-z_{1}}{z-z_{2}}$ , where $R>0$ and $\theta$ is real. c) If $c \neq 0$ and D = 0, prove that f has exactly one fixed point $z_1$ and that it satisfies the equation $\frac{1}{f(z)-z_{1}}=\frac{1}{z-z_{1}}+C$ for some $C \neq 0$ . d) Given any Mobius transformation, investigate the successive images of a given point w. That is, let $w_{1}=f(w), \quad w_{2}=f\left(w_{1}\right), \quad \ldots, \quad w_{n}=f\left(w_{n-1}\right), \quad \ldots$ , and study the behavior of the sequence $\left\{w_{n}\right\}$ . Consider the special case a, b, c, d real, $ad - be = 1$ .