$$ \lim_{x\rarr3} $$$$ \frac{6x-3}{2x-4} $$

To solve the limit

\[
\lim_{x \to 3} \frac{6x - 3}{2x - 4}
\]

we can start by directly substituting $ x = 3 $ into the function:

\[
\frac{6(3) - 3}{2(3) - 4} = \frac{18 - 3}{6 - 4} = \frac{15}{2}
\]

Thus, the limit is:

\[
\lim_{x \to 3} \frac{6x - 3}{2x - 4} = \frac{15}{2}
\]

Let me know if you would like more details or have any questions!

Here are 5 related questions:
1. How do you solve limits involving indeterminate forms like $ \frac{0}{0} $?
2. What happens if direct substitution in a limit leads to infinity?
3. How do you calculate limits approaching infinity?
4. How can you apply L'Hôpital's Rule to find limits?
5. What is the definition of continuity in terms of limits?

**Tip:** Always check if a limit can be solved by direct substitution first before applying more advanced techniques!

$$ \lim_{x \to 3} \frac{6x - 3}{2x - 4} $$

Learn how to solve the limit of the rational function $ \frac{6x - 3}{2x - 4} $ as x approaches 3 using direct substitution. This detailed solution demonstrates key limit concepts and techniques. Suitable for students in Grades 10-12.

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