\lim _{x\to }\left(2\right)\:\frac{3x-2}{x^2+1}

To find the limit 

\[
\lim _{x \to 2}\frac{3x-2}{x^2+1}
\]

we'll substitute \( x = 2 \) into the function and evaluate:

\[
\frac{3(2) - 2}{2^2 + 1}
\]

Now, calculate each term:

1. \( 3(2) = 6 \), so \( 3x - 2 = 6 - 2 = 4 \).
2. \( 2^2 = 4 \), so \( x^2 + 1 = 4 + 1 = 5 \).

Thus, the expression simplifies to:

\[
\frac{4}{5}
\]

So, the limit is:

\[
\lim _{x \to 2}\frac{3x-2}{x^2+1} = \frac{4}{5}
\]

### Answer:
\[
\lim _{x \to 2}\frac{3x-2}{x^2+1} = \frac{4}{5}
\]

Would you like more details or have any questions?

### Related Questions:

1. What is the limit of \(\frac{3x - 2}{x^2 + 1}\) as \(x \to \infty\)?
2. How do you compute limits involving trigonometric functions?
3. What is L'Hopital's Rule and when do you use it?
4. How do you determine if a function is continuous at a point?
5. What is the difference between one-sided and two-sided limits?

### Tip:
When evaluating a limit, always start by substituting the value directly. If direct substitution leads to an indeterminate form (like \( \frac{0}{0} \)), consider algebraic manipulation or using L'Hopital's Rule.

Discover how to solve the limit of the expression (3x - 2) / (x^2 + 1) as x approaches 2 using the substitution method. This calculus problem involves evaluating limits by direct substitution, suitable for students in Grades 10-12.

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