Q. 2 TB - Calculus

Question

Find each of the following limits.

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

Step-by-Step Solution

Answer

Limit values are the followings.

$\begin{array}{l}\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}=-1\\ \underset{h\to 0}{\lim }\frac{\left(3(-1+h{)}^2+1\right)-4}{h}=-6\\ \underset{h\to 0}{\lim }\frac{\frac{1}{2+h}-0.5}{h}=\frac{-1}{4}\\ \underset{z\to 2}{\lim }\frac{z^2-4}{z-2}=4\\ \underset{z\to 4}{\lim }\frac{(1-3z)+11}{z-4}=-3\\ \underset{z\to 1}{\lim }\frac{\frac{1}{z}-1}{z-1}=-1\end{array}$

Step 1. GIven information.

The given limits are following.

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

Step 2. Limit value of lim h → 0 ( 1 - h ) - 1 h

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

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

Step 3. Limit value of lim h → 0 3 ( - 1 + h ) 2 + 1 - 4 h

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

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

Step 4. Limit value of lim h → 0 1 2 + h - 0 . 5 h

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

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

Step 5. Limit value of lim z → 2 z 2 - 4 z - 2

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

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

Step 6. Limit value of lim z → 4 ( 1 - 3 z ) + 11 z - 4

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

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

Step 7. Limit value of lim z → 1 1 z - 1 z - 1

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

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