Chapter 12, Problem 52RE

To determine

Find the limit of the function from calculus.

Answer to Problem 52RE

The limit of the function is $\boxed{2x-5}$

Explanation of Solution

Given information:

Find the limit of the function.

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

Where,

  $f\left(x\right)=x^2-5x-2$

Calculation:

Consider the function,

  $f\left(x\right)=x^2-5x-2$

Direct substitution leads the indeterminate form $\frac{0}{0}$

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

Now consider the limit

  $\underset{h\to 0}{\lim }\frac{2xh+h^2-5h}{h}$

Determine the limit by factorisation,

  $\underset{h\to 0}{\lim }\frac{2xh+h^2-5h}{h}=\underset{h\to 0}{\lim }\frac{\cancel{h}\left(2x+h-5\right)}{\cancel{h}}$

Now lead $h\to 0$ so $h\ne 0$

  $\begin{array}{l}\underset{h\to 0}{\lim }\left(2x+h-5\right)\\ =\underset{h\to 0}{\lim }\left(2x+0-5\right)\\ =2x-5\end{array}$

  $\boxed{2x-5}$

Hence the limit of the function is $\boxed{2x-5}$