Chapter 23, Problem 1RE

To determine

Whether the statement “$\underset{x\to 3}{\lim }\frac{x^2-3x}{x-3}=0$” is true or false..

Answer to Problem 1RE

The given statement is $\underset{\_}{\text{false}}$.

Explanation of Solution

Formula used:

The limit of a function $f\left(x\right)$ is that value of the limit of the function approaches as x approaches the given value a and it is written as $\underset{x\to a}{\lim }f\left(x\right)=f\left(a\right)$.

Calculation:

The given limit is $\underset{x\to 3}{\lim }\frac{x^2-3x}{x-3}=0$.

Evaluate the given limit as as shown below:

$\begin{array}{c}\underset{x\to 3}{\lim }\frac{x^2-3x}{x-3}=\underset{x\to 3}{\lim }\frac{x\left(x-3\right)}{x-3}\\ =\underset{x\to 3}{\lim }x\\ =3\end{array}$

Thus, $\text{LHS}\ne \text{RHS}$.

That is, the value of the given limit is obtained as 3 and hence the given statement is $\underset{\_}{\text{false}}$.