Chapter 11, Problem 50RE

To determine

To calculate:

The value of $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ for the following functions:

  $f\left(x\right)=x^2-5x-2$

Answer to Problem 50RE

The value of $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ for the function $f\left(x\right)=x^2-5x-2$ is $2x-5$ .

Explanation of Solution

Given information:

  $f\left(x\right)=x^2-5x-2$

Calculation:

The given function is:

  $f\left(x\right)=x^2-5x-2$

Need to find $\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ of $f\left(x\right)=x^2-5x-2$ as:

  $\begin{array}{l}\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{\left[{\left(x+h\right)}^2-5\left(x+h\right)-2\right]-\left(x^2-5x-2\right)}{h}\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{x^2+2xh+h^2-5x-5h-2-x^2+5x+2}{h}\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{2xh+h^2-5h}{h}\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\frac{h\left(2x+h-5\right)}{h}\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{h\to 0}{\lim }\left(2x+h-5\right)\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=2x+\left(0\right)-5\left[\because h=0\right]\\ \underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=2x-5\end{array}$