Chapter 2.3, Problem 9E

To determine

To evaluate: The limit of the function $\underset{x\to 5}{\lim }\frac{x^2-6x+5}{x-5}$.

Answer to Problem 9E

The limit of the function is 4.

Explanation of Solution

Limit Laws:

Suppose that c is a constant and the limits $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exists, then

Limit law 2: $\underset{x\to a}{\lim }\left[f\left(x\right)-g\left(x\right)\right]=\underset{x\to a}{\lim }f\left(x\right)-\underset{x\to a}{\lim }g\left(x\right)$

Limit law 7: $\underset{x\to a}{\lim }c=c$

Limit law 8: $\underset{x\to a}{\lim }x=a$

Direct substitution property:

If f is a polynomial or a rational function and a is in the domain of f, then $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$.

Fact 1:

If $f\left(x\right)=g\left(x\right)$ when $x\ne a$, then $\underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a}{\lim }g\left(x\right)$, provided the limit exist.

Calculation:

Compute the limit value of the denominator.

$\begin{array}{c}\underset{x\to 5}{\lim }\left(x-5\right)=\underset{x\to 5}{\lim }\left(x\right)-\underset{x\to 5}{\lim }\left(5\right)\left(\text{by limit law 2}\right)\\ =\underset{x\to 5}{\lim }\left(x\right)-5\left(\text{by limit law 7}\right)\\ =5-5\left(\text{by limit law 8}\right)\\ =0\end{array}$

Since the limit of the denominator is zero, the quotient law cannot be applied.

Let $f\left(x\right)=\frac{x^2-6x+5}{x-5}$ (1)

Simplify $f\left(x\right)$ by using elementary algebra.

Factorise the numerator $x^2-6x+5$.

$\begin{array}{c}{x}^2-6x+5=x^2-5x-1x+5\\ =x\left(x-5\right)-\left(x-5\right)\\ =\left(x-5\right)\left(x-1\right)\end{array}$

Substitute $\left(x-5\right)\left(x-1\right)$ for $x^2-6x+5$ in equation (1).

$f\left(x\right)=\frac{\left(x-5\right)\left(x-1\right)}{x-5}$

Cancel the common terms from both the numerator and the denominator, $f\left(x\right)=\left(x-1\right)$.

Use fact 1, $f\left(x\right)=\left(x-1\right)$ and $x\ne 5$, then $\underset{x\to 5}{\lim }\frac{x^2-6x+5}{x-5}=\underset{x\to 5}{\lim }\left(x-1\right)$.

Apply direct substitution property on the limit function.

$\begin{array}{c}\underset{x\to 5}{\lim }\left(x-1\right)=5-1\\ =4\end{array}$

Thus, the limit of the function is 4.