To solve the limit
[tex]\[
\lim _{x \rightarrow 1} \frac{\sqrt{3 x+1}-2}{x-1}
\][/tex]
we need to handle the indeterminate form that occurs when directly substituting [tex]\( x = 1 \)[/tex]. Here’s a step-by-step detailed way to approach this limit:
1. Substitution:
Substitute [tex]\( x = 1 \)[/tex] directly into the expression:
[tex]\[
\frac{\sqrt{3(1) + 1} - 2}{1 - 1} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}
\][/tex]
This expression is indeterminate, so we need another method to evaluate the limit.
2. Rationalizing the Numerator:
Multiply the numerator and the denominator by the conjugate of the numerator to simplify. The conjugate of [tex]\( \sqrt{3x + 1} - 2 \)[/tex] is [tex]\( \sqrt{3x + 1} + 2 \)[/tex]:
[tex]\[
\frac{\sqrt{3x + 1} - 2}{x - 1} \cdot \frac{\sqrt{3x + 1} + 2}{\sqrt{3x + 1} + 2}
\][/tex]
3. Simplify the Expression:
The numerator becomes a difference of squares:
[tex]\[
\frac{(\sqrt{3x + 1} - 2)(\sqrt{3x + 1} + 2)}{(x - 1)(\sqrt{3x + 1} + 2)} = \frac{(3x + 1) - 4}{(x - 1)(\sqrt{3x + 1} + 2)}
= \frac{3x - 3}{(x - 1)(\sqrt{3x + 1} + 2)}
= \frac{3(x - 1)}{(x - 1)(\sqrt{3x + 1} + 2)}
\][/tex]
4. Cancel Common Factors:
[tex]\[
= \frac{3 \cancel{(x - 1)}}{\cancel{(x - 1)} (\sqrt{3x + 1} + 2)}
= \frac{3}{\sqrt{3x + 1} + 2}
\][/tex]
5. Evaluate the Limit:
As [tex]\( x \)[/tex] approaches 1,
[tex]\[
\lim _{x \rightarrow 1} \frac{3}{\sqrt{3x + 1} + 2} = \frac{3}{\sqrt{3(1) + 1} + 2} = \frac{3}{\sqrt{4} + 2} = \frac{3}{2 + 2} = \frac{3}{4}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim _{x \rightarrow 1} \frac{\sqrt{3x + 1} - 2}{x - 1} = \frac{3}{4}
\][/tex]
