Question

\\\lim_{x \to 3} \frac{2x}{x-3}\\

Explanation:

Identify the limit expression

We analyze the given limit:

$$ \lim_{x \to 3} \frac{2x}{x-3} $$

We need to evaluate the behavior of the function as \(x\) approaches \(3\).

Evaluate direct substitution

We substitute \(x = 3\) into the numerator and denominator:

• Numerator: \(2(3) = 6\)
• Denominator: \(3 - 3 = 0\)

Since the limit is of the form \(\frac{c}{0}\) where \(c
eq 0\), the limit does not exist as a finite real number and represents an infinite limit.

Analyze the left-hand limit

We examine the limit as \(x\) approaches \(3\) from the left (\(x \to 3^-\)):

• For \(x < 3\), the term \(x - 3 < 0\).
• The numerator \(2x\) remains positive (close to \(6\)).
• Thus, the ratio is positive divided by negative:

$$ \lim_{x \to 3^-} \frac{2x}{x-3} = -\infty $$

Analyze the right-hand limit

We examine the limit as \(x\) approaches \(3\) from the right (\(x \to 3^+\)):

• For \(x > 3\), the term \(x - 3 > 0\).
• The numerator \(2x\) remains positive (close to \(6\)).
• Thus, the ratio is positive divided by positive:

$$ \lim_{x \to 3^+} \frac{2x}{x-3} = +\infty $$

Determine the overall limit

Since the left-hand limit and the right-hand limit do not agree:

$$ \lim_{x \to 3^-} \frac{2x}{x-3} eq \lim_{x \to 3^+} \frac{2x}{x-3} $$

The overall limit does not exist (DNE).

Answer:

The limit does not exist (DNE) because the one-sided limits are unequal:

$$ \lim_{x \to 3^-} \frac{2x}{x-3} = -\infty \quad \text{and} \quad \lim_{x \to 3^+} \frac{2x}{x-3} = +\infty $$