Determine whether the statement is true or false. \[ \lim _{x \rightarrow 3}\left(\frac{2 x}{x-3}-\frac{6}{x-3}\right)=\lim _{x \rightarrow 3} \frac{2 x}{x-3}-\lim _{x \rightarrow 3} \frac{6}{x-3} \] True False

To determine whether the statement is true or false, we need to analyze both sides of the equation. The left-hand side is: \[ \lim _{x \rightarrow 3}\left(\frac{2 x}{x-3}-\frac{6}{x-3}\right) \] We can combine the fractions: \[ \lim _{x \rightarrow 3}\left(\frac{2x - 6}{x-3}\right) \] Now, we can factor the numerator: \[ 2x - 6 = 2(x - 3) \] Thus, we have: \[ \lim _{x \rightarrow 3}\left(\frac{2(x - 3)}{x - 3}\right) \] For \(x \neq 3\), we can simplify this to: \[ \lim _{x \rightarrow 3} 2 = 2 \] Now, let's evaluate the right-hand side: \[ \lim _{x \rightarrow 3} \frac{2 x}{x-3} - \lim _{x \rightarrow 3} \frac{6}{x-3} \] Both limits on the right-hand side approach infinity as \(x\) approaches 3, since the denominators approach zero. Therefore, we have: \[ \lim _{x \rightarrow 3} \frac{2 x}{x-3} = \infty \quad \text{and} \quad \lim _{x \rightarrow 3} \frac{6}{x-3} = \infty \] Thus, the right-hand side becomes: \[ \infty - \infty \] which is an indeterminate form. Since the left-hand side evaluates to 2 and the right-hand side is indeterminate, we conclude that the statement is: **False**.