Question

$\lim_{x \to 3}\frac{\sqrt{3x-3}}{\sqrt{2x-4}-\sqrt{2}}$ is equal to

• A. $\sqrt{3}$
• B. $\sqrt{6}$
• C. $\frac{\sqrt{6}}{2}$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

Answer 2

The given limit is $\lim_{x \to 3}\frac{\sqrt{3x-3}}{\sqrt{2x-4}-\sqrt{2}}$.

Upon direct substitution of $x=3$: Numerator: $\sqrt{3(3)-3} = \sqrt{9-3} = \sqrt{6}$ Denominator: $\sqrt{2(3)-4} - \sqrt{2} = \sqrt{6-4} - \sqrt{2} = \sqrt{2} - \sqrt{2} = 0$

The form of the limit is $\frac{\sqrt{6}}{0}$. This indicates that the limit does not exist as a finite value. To determine the behavior, we analyze the one-sided limits:

Right-hand limit ($x \to 3^+$): As $x$ approaches 3 from the right ($x > 3$), $2x-4$ approaches $2$ from values greater than $2$. Thus, $\sqrt{2x-4}$ approaches $\sqrt{2}$ from values greater than $\sqrt{2}$. The denominator $\sqrt{2x-4} - \sqrt{2}$ approaches $0$ from the positive side ($0^+$). The numerator approaches $\sqrt{6}$ (a positive value). So, $\lim_{x \to 3^+}\frac{\sqrt{3x-3}}{\sqrt{2x-4}-\sqrt{2}} = \frac{\sqrt{6}}{0^+} = +\infty$.

Left-hand limit ($x \to 3^-$): As $x$ approaches 3 from the left ($x < 3$), $2x-4$ approaches $2$ from values less than $2$ (but greater than $-4$ for the square root to be defined). Thus, $\sqrt{2x-4}$ approaches $\sqrt{2}$ from values less than $\sqrt{2}$. The denominator $\sqrt{2x-4} - \sqrt{2}$ approaches $0$ from the negative side ($0^-$). The numerator approaches $\sqrt{6}$ (a positive value). So, $\lim_{x \to 3^-}\frac{\sqrt{3x-3}}{\sqrt{2x-4}-\sqrt{2}} = \frac{\sqrt{6}}{0^-} = -\infty$.

Since the left-hand limit ($-\infty$) and the right-hand limit ($+\infty$) are not equal, the limit $\lim_{x \to 3}\frac{\sqrt{3x-3}}{\sqrt{2x-4}-\sqrt{2}}$ does not exist.