Chapter 12.2, Problem 6E

To determine

To find: The way to evaluate the limit of a difference quotient and how to evaluate a limit using the squeeze theorem.

Answer to Problem 6E

The way to evaluate the limit of a difference quotient and how to evaluate a limit using the squeeze theorem is described.

Explanation of Solution

Consider the direct substitution into the difference quotient always produces the intermediate form $\frac{0}{0}$ .

Consider the example for the limit is,

  $\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}$

Solve the above function for the function $f\left(x\right)=3x-1$ .

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0}{\lim }\frac{\left[3\left(2+h\right)-1\right]-\left(3\cdot 2-1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{3h}{h}\\ =3\end{array}$

The squeeze theorem states that the for $h\left(x\right)\le f\left(x\right)\le g\left(x\right)$ is true for all the value of $x$ in an open interval that contains $c$ except possibly at $c$ itself and if,

  $\begin{array}{c}\underset{x\to c}{\text{lim}}h\left(x\right)=L\\ =\underset{x\to c}{\text{lim}}g\left(x\right)\end{array}$

The above equation shows that the limit $\underset{x\to c}{\text{lim}}f\left(x\right)$ exists and is equal to $L$ .