Q. 2TB - Calculus

Question

Limit calculations: Find each of the following limits.

$\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{2+h}}-0.5}{h}$

$\underset{z\to 2}{\lim }\frac{z^2-4}{z-2}$

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

$\underset{z\to 1}{\lim }\frac{{\displaystyle \frac{1}{z}}-1}{z-1}$

Step-by-Step Solution

Answer

The limit values are:

Part(a) Step1.Given Information

Given limit :$\underset{h\to 0}{\lim }\frac{(1-h)-1}{h}$ .

Step2.The solution

The solution $$\begin{gathered} =\underset{h\to 0}{\lim }\frac{(1-h)-1}{h} \\ =\underset{h\to 0}{\lim }\frac{\cancel{1}-h\cancel{-1}}{h} \\ =\underset{h\to 0}{\lim }\frac{-\cancel{h}}{\cancel{h}} \\ =-1 . \end{gathered}$$

Part(b) Step1.Given information

Given limit:$\underset{h\to 0}{\lim }\frac{\left(3\right(-1+h^2)+1)-4}{h}$ .

Step 2.The solution

The solution:$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{(3{(-1+h)}^2+1)-4}{h} \\ =\underset{h\to 0}{\lim }\frac{\left(3\right(h^2-2h+1)+1)-4}{h} \\ =\underset{h\to 0}{\lim }\frac{3h^2-6h+3+1-4}{h} \\ =\underset{h\to 0}{\lim }\frac{3h^2-6h+\cancel{4}-\cancel{4}}{h} \\ =\underset{h\to 0}{\lim }\frac{\cancel{h}(3h-6)}{\cancel{h}} \\ =3\left(0\right)-6 \\ =-6. \end{gathered}$$