Let's evaluate the limit:
[tex]\[
\lim_{x \to \frac{3}{2}} \frac{4x^2 - 9}{2x + 3}
\][/tex]
To solve this, follow these steps:
1. Substitute [tex]\( x = \frac{3}{2} \)[/tex] into the function:
Substitute directly into the expression to see if it results in an indeterminate form (such as [tex]\( \frac{0}{0} \)[/tex]):
[tex]\[
\frac{4 \left( \frac{3}{2} \right)^2 - 9}{2 \left( \frac{3}{2} \right) + 3}
\][/tex]
Compute the numerator and the denominator separately:
[tex]\[
4 \left( \frac{3}{2} \right)^2 = 4 \cdot \frac{9}{4} = 9
\][/tex]
[tex]\[
2 \left( \frac{3}{2} \right) + 3 = 3 + 3 = 6
\][/tex]
Substituting these into the function:
[tex]\[
\frac{9 - 9}{6} = \frac{0}{6} = 0
\][/tex]
Thus, the function's value at [tex]\( x = \frac{3}{2} \)[/tex] is [tex]\( 0 \)[/tex].
So, the solution to the problem is:
[tex]\[
\lim_{x \to \frac{3}{2}} \frac{4x^2 - 9}{2x + 3} = 0
\][/tex]
