Chapter 13.2, Problem 17E

To determine

To evaluate the limit it exist

Answer to Problem 17E

The solution is 12

Explanation of Solution

Given information:

Given equation is,

  $\underset{x\to 0}{\lim }\frac{{\left(2+h\right)}^3-8}{h}$

CALCULATION:

To find the limit,

  $\underset{x\to a}{\lim f\left(x\right)}=L$

“The limit of $f\left(x\right)$ as $x$ approaches $+a$ , equals $L\text{'}$ .

The limit is obtained as:

  $\begin{array}{l}\underset{x\to 0}{\lim }\frac{{\left(2+h\right)}^3-8}{h}\\ =\underset{x\to 0}{\lim }\frac{8+h^3+6h^2+12h-8}{h}\left[\text{apply }{\left(a+b\right)}^2=a^3+b^3+3a^{2}b+3a{b}^2\right]\\ =\underset{x\to 0}{\lim }\frac{h\left(h^2+bh+12\right)}{h}\left[\text{expand}\right]\\ =\underset{x\to 0}{\lim }\left(b^2+6b+12\right)\left[\text{simplify}\right]\\ =0+0+12\left[\text{substitute,}b=0\right]\\ =12\end{array}$

Hence,

The solution is 12