Question

1. \\(\lim_{x \to 1} \frac{\sqrt{3x + 1} - 2}{x - 1}\\)

Explanation:

Identify the indeterminate form

We evaluate the limit directly.

$$ \lim_{x \to 1} \frac{\sqrt{3x + 1} - 2}{x - 1} $$

Substituting \(x = 1\) gives:

$$ \frac{\sqrt{3(1) + 1} - 2}{1 - 1} = \frac{2 - 2}{0} = \frac{0}{0} $$

This is an indeterminate form.

Rationalize the numerator

Multiply by the conjugate expression.

$$ \frac{\sqrt{3x + 1} - 2}{x - 1} \cdot \frac{\sqrt{3x + 1} + 2}{\sqrt{3x + 1} + 2} $$

Simplify the numerator using difference of squares.

$$ \frac{(3x + 1) - 4}{(x - 1)(\sqrt{3x + 1} + 2)} = \frac{3x - 3}{(x - 1)(\sqrt{3x + 1} + 2)} $$

Simplify the expression

Factor the numerator to cancel terms.

$$ \frac{3(x - 1)}{(x - 1)(\sqrt{3x + 1} + 2)} $$

For \(x
eq 1\), cancel \(x - 1\).

$$ \frac{3}{\sqrt{3x + 1} + 2} $$

Evaluate the limit

Substitute \(x = 1\) into simplified form.

$$ \lim_{x \to 1} \frac{3}{\sqrt{3x + 1} + 2} = \frac{3}{\sqrt{4} + 2} = \frac{3}{4} $$

Answer:

\(\frac{3}{4}\)