$$ \lim_{X\text{ }2\text{/}3} $$$$ \frac{2X+3}{X-1} $$$$ = $$

To solve the limit $ \lim_{X \to \frac{2}{3}} \frac{2X + 3}{X - 1} $, we substitute $ X = \frac{2}{3} $ into the given rational function $ \frac{2X + 3}{X - 1} $.

### Step-by-step solution:

1. Substituting $ X = \frac{2}{3} $ into the numerator and denominator:
   \[
   \frac{2 \left(\frac{2}{3}\right) + 3}{\frac{2}{3} - 1}
   \]

2. Simplify the numerator:
   \[
   2 \times \frac{2}{3} = \frac{4}{3}
   \]
   So, the numerator becomes:
   \[
   \frac{4}{3} + 3 = \frac{4}{3} + \frac{9}{3} = \frac{13}{3}
   \]

3. Simplify the denominator:
   \[
   \frac{2}{3} - 1 = \frac{2}{3} - \frac{3}{3} = \frac{-1}{3}
   \]

4. The expression now becomes:
   \[
   \frac{\frac{13}{3}}{\frac{-1}{3}} = \frac{13}{3} \times \frac{3}{-1} = -13
   \]

Thus, the limit is:
\[
\lim_{X \to \frac{2}{3}} \frac{2X + 3}{X - 1} = -13
\]

Would you like more details, or do you have any questions?

Here are 5 related questions to explore:

1. What happens when $ X $ approaches 1 instead of $ \frac{2}{3} $?
2. How does the limit change if the expression is $ \frac{2X - 3}{X + 1} $?
3. Can we apply L'Hopital's Rule to this type of limit?
4. What if the limit approaches infinity instead of a specific value?
5. How do we handle limits involving indeterminate forms like $ \frac{0}{0} $?

**Tip:** Always check if you can simplify a function before applying limit rules!

\lim_{X \to \frac{2}{3}} \frac{2X+3}{X-1}

Solve the limit $ \lim_{X \to \frac{2}{3}} \frac{2X+3}{X-1} $ with a detailed step-by-step process. Learn key mathematical concepts such as substitution in limits and simplifying rational functions. Suitable for Grades 10-12.

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