Find each of the following limits.

$\begin{array}{l}\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}\\ \underset{h\to 0}{\lim }\frac{\left(3(-1+h{)}^2+1\right)-4}{h}\\ \underset{h\to 0}{\lim }\frac{\frac{1}{2+h}-0.5}{h}\\ \underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}\\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}\end{array}$

The given limits are following.

$\begin{array}{l}\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}\\ \underset{h\to 0}{\lim }\frac{\left(3(-1+h{)}^2+1\right)-4}{h}\\ \underset{h\to 0}{\lim }\frac{\frac{1}{2+h}-0.5}{h}\\ \underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}\\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}\end{array}$

Determine the limit $\begin{array}{l}\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}\end{array}.$

$$\begin{gathered} \begin{array}{l}\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}\end{array}=\begin{array}{l}\underset{h\to 0}{\lim }\frac{-h}{h}\end{array} \\ \underset{h\to 0}{\lim }\frac{(1-h)-1}{h}=\begin{array}{l}\underset{h\to 0}{\lim }-1\end{array} \\ \underset{h\to 0}{\lim }\frac{(1-h)-1}{h}=-1 \end{gathered}$$

Determine the limit $\begin{array}{l}\underset{h\to 0}{\lim }\frac{\left(3(-1+h{)}^2+1\right)-4}{h}\end{array}.$

$$\begin{gathered} \begin{array}{l}\underset{h\to 0}{\lim }\frac{\left(3(-1+h{)}^2+1\right)-4}{h}\end{array}=\begin{array}{l}\underset{h\to 0}{\lim }\frac{\left(3(1-2h+h{)}^2+1\right)-4}{h}\end{array} \\ =\begin{array}{l}\underset{h\to 0}{\lim }\frac{3-6h+3h^2+1-4}{h}\end{array} \\ =\begin{array}{l}\underset{h\to 0}{\lim }\frac{h(3h-6)}{h}\end{array} \\ =-6 \end{gathered}$$

Determine the limit $\begin{array}{l}\underset{h\to 0}{\lim }\frac{\frac{1}{2+h}-0.5}{h}\end{array}.$

$\begin{array}{rl}& \underset{h\to 0}{lim}\frac{\frac{1}{2+h}-0.5}{h}=\underset{h\to 0}{lim}\frac{\frac{1-1-0.5h}{2+h}}{h}\\ & \underset{h\to 0}{lim}\frac{\frac{1}{2+h}-0.5}{h}=\underset{h\to 0}{lim}\frac{-0.5h}{2+h}\times \frac{1}{h}\\ & \underset{h\to 0}{lim}\frac{\frac{1}{2+h}-0.5}{h}=\frac{-1}{4}\end{array}$

Determine the limit $\begin{array}{l}\underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\end{array}.$

$$\begin{gathered} \begin{array}{l}\underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\end{array}=\underset{z\to 2}{\lim }\frac{(z-2)(z+2)}{z-2} \\ \begin{array}{l}\underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\end{array}=\begin{array}{l}\underset{z\to 2}{\lim }(z+2)\end{array} \\ \begin{array}{l}\underset{z\to 2}{\lim }\frac{z^2-4}{z-2}\end{array}=4 \end{gathered}$$

Determine the limit $\underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}.$

$$\begin{gathered} \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}=\underset{z\to 4}{lim}\frac{-3z+12}{z-4} \\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}=\underset{z\to 4}{lim}\frac{-3(z-4)}{z-4} \\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}=\underset{z\to 4}{lim}\left(-3\right) \\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}=-3 \end{gathered}$$

Determine the limit $\underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}.$

$$\begin{gathered} \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}=\underset{z\to 1}{lim}\frac{\frac{1-z}{z}}{z-1} \\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}=\underset{z\to 1}{lim}\left(\frac{-(z-1)}{z}\times \frac{1}{z-1}\right) \\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}=\underset{z\to 1}{lim}\left(\frac{-1}{z}\right) \\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}=-1 \end{gathered}$$