Chapter 15, Problem 21SGA

To determine

To evaluate: The given limit.

Answer to Problem 21SGA

  $\frac{1}{5}$

Explanation of Solution

Given:

  $\underset{x\to 5}{\lim }\frac{x^2-9x+20}{x^2-5x}$

Formula Used:

  $\begin{array}{l}\underset{x\to a}{\lim }\left[f\left(x\right)+g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)\\ \underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)\\ \underset{x\to a}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\frac{\underset{x\to a}{\lim }f\left(x\right)}{\underset{x\to a}{\lim }g\left(x\right)}\end{array}$

Calculation:

  $\begin{array}{l}\underset{x\to 5}{\lim }\frac{x^2-9x+20}{x^2-5x}\\ \Rightarrow \underset{x\to 5}{\lim }\frac{x^2-4x-5x+20}{x\left(x-5\right)}\\ \Rightarrow \underset{x\to 5}{\lim }\frac{\left(x-4\right)\left(x-5\right)}{x\left(x-5\right)}\\ \Rightarrow \frac{\underset{x\to 5}{\lim }\left(x-4\right)}{\underset{x\to 5}{\lim }x}\\ \Rightarrow \frac{5-4}{5}\\ \Rightarrow \frac{1}{5}\end{array}$

Conclusion:

Thus the solution of the given function is $\frac{1}{5}$