Calculus: Early Transcendentals 8th Edition

$$\lim_{x\to1/2}\frac{6x^2+5x-4}{4x^2+16x-9}=\frac{11}{20}$$
$$A=\lim_{x\to1/2}\frac{6x^2+5x-4}{4x^2+16x-9}$$ Again, the 2 methods are both applicable here. 1) Method 1: Elementary method $$A=\lim_{x\to1/2}\frac{(6x^2-3x)+(8x-4)}{(4x^2-2x)+(18x-9)}$$ $$A=\lim_{x\to1/2}\frac{3x(2x-1)+4(2x-1)}{2x(2x-1)+9(2x-1)}$$ $$A=\lim_{x\to1/2}\frac{(2x-1)(3x+4)}{(2x-1)(2x+9)}$$ $$A=\lim_{x\to1/2}\frac{3x+4}{2x+9}$$ $$A=\frac{3\times\frac{1}{2}+4}{2\times\frac{1}{2}+9}$$ $$A=\frac{\frac{11}{2}}{10}=\frac{11}{20}$$ 2) Method 2: L'Hospital's Rule Because $$\lim_{x\to1/2}(6x^2+5x-4)=6\times(\frac{1}{2})^2+5\times\frac{1}{2}-4=\frac{3}{2}+\frac{5}{2}-4=\frac{8}{2}-4=0$$ and $$\lim_{x\to1/2}(4x^2+16x-9)=4(\frac{1}{2})^2+16\times\frac{1}{2}-9=1+8-9=0,$$ this limit is an indeterminate form of $\frac{0}{0}$. Therefore, we can apply L'Hospital's Rule: $$A=\lim_{x\to1/2}\frac{\frac{d}{dx}(6x^2+5x-4)}{\frac{d}{dx}(4x^2+16x-9)}$$ $$A=\lim_{x\to1/2}\frac{12x+5}{8x+16}$$ $$A=\frac{12\times\frac{1}{2}+5}{8\times\frac{1}{2}+16}$$ $$A=\frac{6+5}{4+16}=\frac{11}{20}$$