Chapter 2.3, Problem 11E

To determine

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

Answer to Problem 11E

The limit of the function does not exist.

Explanation of Solution

Limit Laws:

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

Reason:

The limit of the denominator is zero.

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

The limit of the numerator is 6.

$\begin{array}{c}\underset{x\to 5}{\lim }\left(x^2-5x+6\right)=\underset{x\to 5}{\lim }\left(x^2\right)-\underset{x\to 5}{\lim }\left(5x\right)+\underset{x\to 5}{\lim }\left(6\right)\left(\text{by limit law 2 and 1}\right)\\ =\underset{x\to 5}{\lim }\left(x^2\right)-5\underset{x\to 5}{\lim }\left(x\right)+\underset{x\to 5}{\lim }\left(6\right)\left(\text{by limit law 3}\right)\\ ={\left(5\right)}^2-5\left(5\right)+6\left(\text{by limit law 9,8 and 7}\right)\\ =6\end{array}$

The quotient law cannot be used as the denominator of the function tends to zero as $x\to 5$.

So, $\underset{x\to 5}{\lim }\frac{x^2-5x+6}{x-5}$ does not exist as the numerator of the function approaches 6 when x tends to 5 and the denominator of the function approaches 0 when x tends to 5.

Thus, the limit of the function does not exist.