The problem is to evaluate the limit:

$\lim_{x \to 10} \sqrt{3x + 1}$

Step-by-step solution:

1. Direct Substitution: Since there are no indeterminate forms (like $\frac{0}{0}$ or $\infty - \infty$), we can directly substitute $x = 10$ into the expression $\sqrt{3x + 1}$.

   $\lim_{x \to 10} \sqrt{3x + 1} = \sqrt{3(10) + 1}$

2. Simplify the expression: $\sqrt{3(10) + 1} = \sqrt{30 + 1} = \sqrt{31}$

So, the value of the limit is:

$\boxed{\sqrt{31}}$

Would you like more details on the solution or related concepts?

Related questions:

1. What is the difference between limits and continuity?
2. How do we handle limits that result in indeterminate forms?
3. What is the process for solving limits involving trigonometric functions?
4. Can limits always be solved using direct substitution?
5. How do limits apply to the definition of derivatives in calculus?

Tip:

When solving limits, always try direct substitution first. If it results in an indeterminate form, explore algebraic simplification, factoring, or using L'Hopital's Rule for more complex cases.