Answer
$\dfrac{7}{4}$
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to \frac{1}{2}} \dfrac{x^2+3x}{2x^2+3x-1}$.
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to \frac{1}{2}} \dfrac{x^2+3x}{2x^2+3x-1}=\dfrac{(1/2)^2+3\times (1/2)}{2(1/2)^2+3(1/2)-1}\\=\dfrac{7}{4}$
