Answer
$\dfrac{11}{20}$
Work Step by Step
Given: $\lim\limits_{x\to1/2}\dfrac{6x^2+5x-4}{4x^2+16x-9}$
This shows an indeterminate form of type $\dfrac{0}{0}$, and so we will apply L'Hospital Rule.
$\lim\limits_{x\to1/2}\dfrac{\dfrac{d}{dx}(6x^2+5x-4)}{\dfrac{d}{dx}(4x^2+16x-9)}=\lim\limits_{x\to1/2}\dfrac{12x+5}{8x+16}\\=\dfrac{12\times\dfrac{1}{2}+5}{8\times\dfrac{1}{2}+16}\\=\dfrac{6+5}{4+16}=\dfrac{11}{20}$
